Integration Using Partial Fractions
Partial Fractions
Rational Functions If f(x) and g(x) are polynomial functions such that g(x) ≠ 0 then \(\frac{f(x)}{g(x)}\) is called a rational function.
If degree f(x) < degree g(x) then \(\frac{f(x)}{g(x)}\) is called a proper rational function.
If degree f(x) ≥ degree then \(\frac{f(x)}{g(x)}\) is called an improper rational function.
If \(\frac{f(x)}{g(x)}\) is an improper rational function then by dividing f(x) by g(x), we can express \(\frac{f(x)}{g(x)}\) as the sum of a polynomial and a proper rational function.
Partial Fractions Any proper rational function \(\frac{p(x)}{q(x)}\) can be expressed as the sum of rational functions, each having a simplest factor q(x). Each such fraction is known as a partial fraction and the process of obtaining them is called the decomposition or resolving of the given function into partial fractions.
Read and Learn More Class 12 Math Solutions
Integration By Partial Fractions Formula
Method We first resolve the denominator of the given fraction into simplest factors. Based on these factors, we obtain the corresponding partial fraction as per rules given below:
The values of A, B, C, etc., can be obtained as shown below.
Solved Examples On Partial Fractions
Example 1 Resolve \(\frac{2 x+3}{(x-3)(x+1)}\) into partial fractions.
Solution
Given
\(\frac{2 x+3}{(x-3)(x+1)}\)Let \(\frac{(2 x+3)}{(x-3)(x+1)}=\frac{A}{(x-3)}+\frac{B}{(x+1)} .\)
\(\frac{2 x+3}{(x-3)(x+1)}=\frac{A(x+1)+B(x-3)}{(x-3)(x+1)}\)or (2x + 3) ≡ A(x+1) + B(x-3) …(1)
Putting (x-3) = 0 or x = 3 in (1), we get A = (9/4).
Putting (x+1) = 0 or x = -1 in (1), we get B = (-1/4).
∴ \(\frac{(2 x+3)}{(x-3)(x+1)}=\frac{9}{4(x-3)}-\frac{1}{4(x+1)}\)
Integration By Partial Fractions Formula
Example 2 Resolve \(\frac{x^3-2 x^2-13 x-12}{x^2-3 x-10}\) into partial fractions.
Solution
Given
\(\frac{x^3-2 x^2-13 x-12}{x^2-3 x-10}\)On dividing, we get
\(\frac{x^3-2 x^2-13 x-12}{x^2-3 x-10}=(x+1)-\frac{2}{\left(x^2-3 x-10\right)}\) …(1)
Let \(\frac{2}{\left(x^2-3 x-10\right)}=\frac{2}{(x-5)(x+2)}=\frac{A}{x-5}+\frac{B}{x+2}\)
Then, \(\frac{2}{(x-5)(x+2)}=\frac{A(x+2)+B(x-5)}{(x-5)(x+2)}\)
or 2 ≡ A(x+2) + B(x-5) …(2)
Putting (x-5) = 0 or x = 5 in (2), we get A = (2/7).
Putting (x+2) = 0 or x = -2 in (2), we get B = (-2/7).
∴ \(\frac{2}{\left(x^2-3 x-10\right)}=\frac{2}{7(x-5)}-\frac{2}{7(x+2)}\)
Hence, \(\frac{x^3-2 x^2-13 x-12}{x^2-3 x-10}=(x+1)-\frac{2}{7(x-5)}+\frac{2}{7(x+2)}\).
Understanding Integration by Partial Fractions
Example 3 Resolve \(\frac{16}{(x-2)(x+2)^2}\)
Solution
Given
\(\frac{16}{(x-2)(x+2)^2}\)Let \(\frac{16}{(x-2)(x+2)^2}=\frac{A}{x-2}+\frac{B}{x+2}+\frac{C}{(x+2)^2}\)
or \(\frac{16}{(x-2)(x+2)^2}=\frac{A(x+2)^2+B(x-2)(x+2)+C(x-2)}{(x-2)(x+2)^2}\)
∴ 16 ≡ A(x+2)2 + B(x-2)(x+2) + C(x-2) …(1)
or 16 ≡ (a+b)x2 + (4A+C)x + (4A-4B-2C) …(2)
Putting (x-2) = 0 or x = 2 in (1), we get A = 1.
Putting (x+2) = 0 or x = -2 in (1), we get C = -4.
Comparing coefficients of x2 on both sides of (2), we get
A + B = 0 or B = -A = -1.
Thus A = 1, B = -1 and C = -4.
∴ \(\frac{16}{(x-2)(x+2)^2}=\left[\frac{1}{(x-2)}-\frac{1}{(x+2)}-\frac{4}{(x+2)^2}\right]\)
Different Forms of Partial Fraction Decomposition
Example 4 Resolve \(\frac{2 x+1}{(x-1)\left(x^2+1\right)}\) into partial fractions.
Solution
Given
\(\frac{2 x+1}{(x-1)\left(x^2+1\right)}\)Let \(\frac{2 x+1}{(x-1)\left(x^2+1\right)}=\frac{A}{(x-1)}+\frac{B x+C}{\left(x^2+1\right)}\)
or \(\frac{(2 x+1)}{(x-1)\left(x^2+1\right)}=\frac{A\left(x^2+1\right)+(B x+C)(x-1)}{(x-1)\left(x^2+1\right)} .\)
∴ 2x + 1 ≡ A(x2+1) + (Bx+C)(x-1)
or 2x + 1 ≡ (A+B)x2 + (C-B)x + (A-c) …(1)
Equating the like powers of x on both sides of (1), we get
A + B = 0, C – B = 2 and A – C = 1.
On solving these equations, we get
A = \(\frac{3}{2}\), B = \(-\frac{3}{2}\) and C = \(\frac{1}{2}\).
∴ \(\frac{2 x+1}{(x-1)\left(x^2+1\right)}=\frac{3}{2(x-1)}+\frac{\left(\frac{-3}{2} x+\frac{1}{2}\right)}{x^2+1}=\left[\frac{3}{2(x-1)}+\frac{(1-3 x)}{2\left(x^2+1\right)}\right] .\)
Integration Using Partial Fractions
Integration Using Partial Fractions Solved Examples
Example 1 Evaluate \(\int \frac{(x-1)}{(x+1)(x-2)} d x\).
Solution
Given:
\(\int \frac{(x-1)}{(x+1)(x-2)} d x\)Let \(\frac{(x-1)}{(x+1)(x-2)}=\frac{A}{(x+1)}+\frac{B}{(x-2)} .\) …(1)
Then, (x-1) ≡ A(x-2) + B(x+1)
Putting x = -1 in (1), we get A = \(\frac{2}{3}\).
Putting x = 2 in (1), we get B = \(\frac{1}{3}\).
∴ \(\frac{(x-1)}{(x+1)(x-2)}=\frac{2}{3(x+1)}+\frac{1}{3(x-2)}\)
⇒ \(\int \frac{(x-1)}{(x+1)(x-2)} d x=\frac{2}{3} \int \frac{d x}{(x+1)}+\frac{1}{3} \int \frac{d x}{(x-2)}\)
= \(\frac{2}{3} \log |x+1|+\frac{1}{3} \log |x-2|+C \text {. }\)
\(\int \frac{(x-1)}{(x+1)(x-2)} d x\) = \(\frac{2}{3} \log |x+1|+\frac{1}{3} \log |x-2|+C \text {. }\)
Examples of Partial Fraction Integration
Example 2 Evaluate \(\int \frac{\left(x^2+1\right)}{\left(x^2-5 x+6\right)} d x\).
Solution
Given
\(\int \frac{\left(x^2+1\right)}{\left(x^2-5 x+6\right)} d x\)Here the integrand is not a proper rational function on dividing (x2+1) by (x2-5x+6), we get
\(\frac{\left(x^2+1\right)}{\left(x^2-5 x+6\right)}=1+\frac{(5 x-5)}{\left(x^2-5 x+6\right)}=1+\frac{(5 x-5)}{(x-2)(x-3)} .\)Now, let \(\frac{(5 x-5)}{(x-2)(x-3)}=\frac{A}{(x-2)}+\frac{B}{(x-3)}\)
⇒ \(\frac{(5 x-5)}{(x-2)(x-3)}=\frac{A(x-3)+B(x-2)}{(x-2)(x-3)}\)
⇒ (5x-5) ≡ A(x-3) + B(x-2) …(1)
Putting x = 2 on both sides of (1), we get A = -5.
Putting x = 3 on both sides of (1), we get B = 10.
∴ \(\frac{\left(x^2+1\right)}{\left(x^2-5 x+6\right)}=1-\frac{5}{(x-2)}+\frac{10}{(x-3)}\)
⇒ \(\int \frac{\left(x^2+1\right)}{\left(x^2-5 x+6\right)} d x=\int d x-5 \int \frac{d x}{(x-2)}+10 \int \frac{d x}{(x-3)}\)
= X = 5log |x-2| + 10 log |x-3| + C.
\(\int \frac{\left(x^2+1\right)}{\left(x^2-5 x+6\right)} d x\) = X = 5log |x-2| + 10 log |x-3| + C.
Step-by-Step Guide to Partial Fractions
Example 3 Evaluate \(\int \frac{(3 x-2)}{(x+1)^2(x+3)} d x\)
Solution
Given
\(\int \frac{(3 x-2)}{(x+1)^2(x+3)} d x\)Let \(\frac{(3 x-2)}{(x+1)^2(x+3)}=\frac{A}{(x+1)}+\frac{B}{(x+1)^2}+\frac{C}{(x+3)}\)
⇒ (3x-2) ≡ A(x+1)(x+3) + B(x+3) + C(x+1)2 …(1)
Putting x = -3 on both sides of (1), we get C = \(-\frac{11}{4}\).
Putting x = -1 on both sides of (1), we get B = \(-\frac{5}{2}\).
Comparing coefficients of x2 on both sides of (1), we get
A + C = 0 ⇒ A = -C = \(\frac{11}{4}\).
∴ \(\frac{(3 x-2)}{(x+1)^2(x+3)}=\frac{11}{4(x+1)}-\frac{5}{2(x+1)^2}-\frac{11}{4(x+3)}\)
⇒ \(\int \frac{(3 x-2)}{(x+1)^2(x+3)} d x=\frac{11}{4} \cdot \int \frac{d x}{(x+1)}-\frac{5}{2} \cdot \int \frac{1}{(x+1)^2} d x-\frac{11}{4} \cdot \int \frac{d x}{(x+3)}\)
= \(\frac{11}{4} \cdot \log |x+1|+\frac{5}{2(x+1)}-\frac{11}{4} \cdot \log |x+3|+C\)
= \(\frac{11}{4} \cdot \log \left|\frac{x+1}{x+3}\right|+\frac{5}{2(x+1)}+C\)
\(\int \frac{(3 x-2)}{(x+1)^2(x+3)} d x\) = \(\frac{11}{4} \cdot \log \left|\frac{x+1}{x+3}\right|+\frac{5}{2(x+1)}+C\)
Example 4 Evaluate \(\int \frac{d x}{\left(x^3+x^2+x+1\right)}\).
Solution
Given
\(\int \frac{d x}{\left(x^3+x^2+x+1\right)}\)We have \(\frac{1}{\left(x^3+x^2+x+1\right)}=\frac{1}{x^2(x+1)+(x+1)}=\frac{1}{(x+1)\left(x^2+1\right)} .\)
Let \(\frac{1}{(x+1)\left(x^2+1\right)}=\frac{A}{(x+1)}+\frac{B x+C}{\left(x^2+1\right)}\)
Integration By Partial Fractions Class 12
⇒ 1 ≡ A(x2+1) + (Bx+C)(x+1) …(1)
Putting x = -1 on both sides of (1), we get A = \(\frac{1}{2}\).
Comparing coefficients of x2 on both sides of (1), we get
A + B = 0 ⇒ B = -A = \(-\frac{1}{2}\).
Comparing coefficients of x on both sides of (1), we get
B + C = 0 ⇒ C = -B = \(\frac{1}{2}\).
∴ \(\frac{1}{(x+1)\left(x^2+1\right)}=\frac{1}{2(x+1)}+\frac{\frac{-1}{2} x+\frac{1}{2}}{x^2+1}\)
∴ \(\int \frac{d x}{\left(x^3+x^2+x+1\right)}=\int \frac{d x}{(x+1)\left(x^2+1\right)}\)
= \(\frac{1}{2} \cdot \int \frac{d x}{(x+1)}-\frac{1}{2} \int \frac{x}{\left(x^2+1\right)} d x+\frac{1}{2} \int \frac{d x}{\left(x^2+1\right)}\)
= \(\frac{1}{2} \cdot \int \frac{d x}{(x+1)}-\frac{1}{4} \cdot \int \frac{2 x}{\left(x^2+1\right)} d x+\frac{1}{2} \int \frac{d x}{\left(x^2+1\right)}\)
= \(\frac{1}{2} \log |x+1|-\frac{1}{4} \log \left|x^2+1\right|+\frac{1}{2} \tan ^{-1} x+C .\)
\(\int \frac{d x}{\left(x^3+x^2+x+1\right)}\) = \(\frac{1}{2} \log |x+1|-\frac{1}{4} \log \left|x^2+1\right|+\frac{1}{2} \tan ^{-1} x+C .\)
Common Types of Rational Functions for Partial Fractions
Example 5 Evaluate \(\int \frac{x^4}{(x-1)\left(x^2+1\right)} d x \text {. }\)
Solution
Given
\(\int \frac{x^4}{(x-1)\left(x^2+1\right)} d x \text {. }\) \(\frac{x^4}{(x-1)\left(x^2+1\right)}=\frac{x^4}{\left(x^3-x^2+x-1\right)}=(x+1)+\frac{1}{\left(x^3-x^2+x-1\right)}\)⇒ \(\frac{x^4}{(x-1)\left(x^2+1\right)}=(x+1)+\frac{1}{(x-1)\left(x^2+1\right)}\) …(1)
Let \(\frac{1}{(x-1)\left(x^2+1\right)}=\frac{A}{(x-1)}+\frac{B x+C}{\left(x^2+1\right)}\). Then,
1 ≡ A(x2+1) + (Bx+C)(x-1) …(2)
Putting x = 1 in (2), we get A = \(\frac{1}{2}\).
Comparing coefficients of x2 on both sides of (2), we get
A + B = 0 ⇒ B = -A = \(-\frac{1}{2}\).
Comparing the constant terms on both sides of (2), we get
A – C = 1 ⇒ C = (A-1) (\(\frac{1}{2}\) – 1) = \(-\frac{1}{2}\).
∴ \(\frac{1}{(x-1)\left(x^2+1\right)}=\frac{1}{2(x-1)}+\frac{-\frac{1}{2} x-\frac{1}{2}}{\left(x^2+1\right)}\)
∴ \(\frac{x^4}{(x-1)\left(x^2+1\right)}=(x+1)+\frac{1}{2(x-1)}-\frac{1}{2} \cdot \frac{(x+1)}{\left(x^2+1\right)}\)
Integration By Partial
⇒ \(\int \frac{x^4}{(x-1)\left(x^2+1\right)} d x=\int(x+1) d x+\frac{1}{2} \int \frac{d x}{(x-1)}-\frac{1}{4} \cdot \int \frac{2 x}{\left(x^2+1\right)} d x-\frac{1}{2} \int \frac{d x}{\left(x^2+1\right)}\)
= \(\frac{x^2}{2}+x+\frac{1}{2} \log |x-1|-\frac{1}{4} \log \left(x^2+1\right)-\frac{1}{2} \tan ^{-1} x+C\).
\(\int \frac{x^4}{(x-1)\left(x^2+1\right)} d x \text {. }\) = \(\frac{x^2}{2}+x+\frac{1}{2} \log |x-1|-\frac{1}{4} \log \left(x^2+1\right)-\frac{1}{2} \tan ^{-1} x+C\).
Example 6 Evaluate \(\int \frac{(3 x+5)}{\left(x^3-x^2-x+1\right)} d x\)
Solution
Given
\(\int \frac{(3 x+5)}{\left(x^3-x^2-x+1\right)} d x\)(x3-x2-x+1) = x2(x-1)-(x-1) = (x-1)(x2-1) = (x-1)2(x+1).
Let \(\frac{3 x+5}{\left(x^3-x^2-x+1\right)}=\frac{3 x+5}{(x-1)^2(x+1)}=\frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x+1)}\)
⇒ (3x+5) ≡ A(x-1)(x+1) + B(x+1) + C(x-1)2 …(1)
Putting x = 1 on both sides of (1), we get B = 4.
Putting x = -1 on both sides of (1), we get C = \(\frac{1}{2}\).
Comparing coefficients of x2 on both sides of (1), we get
A + C = 0 ⇒ A = -C = \(-\frac{1}{2}\).
∴ \(\frac{(3 x+5)}{\left(x^3-x^2-x+1\right)}=\frac{-1}{2(x-1)}+\frac{4}{(x-1)^2}+\frac{1}{2(x+1)}\)
⇒ \(\int \frac{(3 x+5)}{\left(x^3-x^2-x+1\right)} d x=-\frac{1}{2} \int \frac{d x}{(x-1)}+4 \int \frac{d x}{(x-1)^2}+\frac{1}{2} \int \frac{d x}{(x+1)}\)
= \(-\frac{1}{2} \log |x-1|-\frac{4}{(x-1)}+\frac{1}{2} \log |x+1|+C .\)
\(\int \frac{(3 x+5)}{\left(x^3-x^2-x+1\right)} d x\) = \(-\frac{1}{2} \log |x-1|-\frac{4}{(x-1)}+\frac{1}{2} \log |x+1|+C .\)
Integration By Partial
Example 7 Evaluate \(\int \frac{\left(x^3-1\right)}{\left(x^3+x\right)} d x .\)
Solution
Given
\(\int \frac{\left(x^3-1\right)}{\left(x^3+x\right)} d x .\)We have
\(\frac{\left(x^3-1\right)}{\left(x^3+x\right)}=1-\frac{(x+1)}{\left(x^3+x\right)}\) [on dividing]
= \(1-\frac{(x+1)}{x\left(x^2+1\right)}\) …(1)
Let \(\frac{(x+1)}{x\left(x^2+1\right)}=\frac{A}{x}+\frac{B x+C}{\left(x^2+1\right)} .\)
Then, (x+1) ≡ A(x2+1) + (Bx+C)x …(2)
Putting x = 0 in (2), we get A = 1.
Comparing coefficients of x in (2), we get C = 1.
Comparing coefficients of x2 in (2), we get
A + B = 0 ⇒ B = -A = -1.
∴ A = 1, B = -1 and C = 1.
Thus, \(\frac{(x+1)}{x\left(x^2+1\right)}=\frac{1}{x}+\frac{(1-x)}{\left(x^2+1\right)}\)
⇒ \(\int \frac{\left(x^3-1\right)}{\left(x^3+x\right)} d x=\int d x-\int \frac{(x+1)}{x\left(x^2+1\right)} d x\)
= \(x-\left\{\int \frac{d x}{x}+\int \frac{(1-x)}{\left(x^2+1\right)} d x\right\}\)
= \(x-\int \frac{d x}{x}-\int \frac{d x}{\left(x^2+1\right)}+\frac{1}{2} \int \frac{2 x}{\left(x^2+1\right)} d x\)
= \(x-\log |x|-\tan ^{-1} x+\frac{1}{2} \log \left(x^2+1\right)+C\).
\(\int \frac{\left(x^3-1\right)}{\left(x^3+x\right)} d x .\) = \(x-\log |x|-\tan ^{-1} x+\frac{1}{2} \log \left(x^2+1\right)+C\).
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Example 8 Evaluate \(\int \frac{\cos x}{(1-\sin x)(2-\sin x)} d x\)
Solution
Given
\(\int \frac{\cos x}{(1-\sin x)(2-\sin x)} d x\)Putting sin x = t and cos x dx = dt, we get
Let \(\frac{1}{(1-t)(2-t)}=\frac{A}{(1-t)}+\frac{B}{(2-t)}\)
⇒ 1 ≡ A(2-t) + B(1-t) …(1)
Putting t = 1 in (1), we get A = 1.
Putting t = 2 in (1), we get B = -1.
∴ \(\frac{1}{(1-t)(2-t)}=\frac{1}{(1-t)}-\frac{1}{(2-t)}\)
⇒ \(\int \frac{\cos x}{(1-\sin x)(2-\sin x)} d x=\int \frac{d t}{(1-t)(2-t)}\)
= \(\int\left\{\frac{1}{(1-t)}-\frac{1}{(2-t)}\right\} d t\)
= \(\int \frac{d t}{(1-t)}-\int \frac{d t}{(2-t)}\)
= -log |1-t| + log |2-t| + C
= \(\log \left|\frac{2-t}{1-t}\right|+C=\log \left|\frac{2-\sin x}{1-\sin x}\right|+C .\)
\(\int \frac{\cos x}{(1-\sin x)(2-\sin x)} d x\) = \(\log \left|\frac{2-t}{1-t}\right|+C=\log \left|\frac{2-\sin x}{1-\sin x}\right|+C .\)
Partial Fractions with Repeated Factors
Example 9 Evaluate \(\int \frac{d x}{x\left\{6(\log x)^2+7 \log x+2\right\}} \text {. }\)
Solution
Given
\(\int \frac{d x}{x\left\{6(\log x)^2+7 \log x+2\right\}} \text {. }\)Putting log x = t and \frac{1}{x} dx = dt, we get
I = \(\int \frac{d x}{x\left\{6(\log x)^2+7 \log x+2\right\}}=\int \frac{d t}{\left(6 t^2+7 t+2\right)}=\int \frac{d t}{(2 t+1)(3 t+2)} .\)
Let \(\frac{1}{(2 t+1)(3 t+2)}=\frac{A}{(2 t+1)}+\frac{B}{(3 t+2)} \text {. }\)
Then, 1 ≡ A(3t+2) + B(2t+1) …(1)
Putting t = \(-\frac{1}{2}\) in (1), we get A = 2.
Putting t = \(-\frac{2}{3}\) in (1), we get B = -3.
∴ \(\frac{1}{(2 t+1)(3 t+2)}=\frac{2}{(2 t+1)}-\frac{3}{(3 t+2)}\)
⇒ I = \(\int \frac{d t}{(2 t+1)(3 t+2)}\)
= \(\int \frac{2 d t}{(2 t+1)}-\int \frac{3 d t}{(3 t+2)}\)
= log |2t+1| – log |3t+2| + C
= \(\log \left|\frac{2 t+1}{3 t+2}\right|+C\)
= \(\log \left|\frac{2 \log x+1}{3 \log x+2}\right|+C\)
\(\int \frac{d x}{x\left\{6(\log x)^2+7 \log x+2\right\}} \text {. }\) = \(\log \left|\frac{2 \log x+1}{3 \log x+2}\right|+C\)
Example 10 Evaluate \(\int \frac{x^2}{\left(1+x^3\right)\left(2+x^3\right)} d x\)
Solution
Given
\(\int \frac{x^2}{\left(1+x^3\right)\left(2+x^3\right)} d x\)Putting x3 = t and x2dx = \(\frac{1}{3} dt\), we get
I = \(\int \frac{x^2}{\left(1+x^3\right)\left(2+x^3\right)} d x=\frac{1}{3} \cdot \int \frac{d t}{(1+t)(2+t)} .\)
Let \(\frac{1}{(1+t)(2+t)}=\frac{A}{(1+t)}+\frac{B}{(2+t)}\). Then,
1 ≡ A(2+t) + B(1+t) …(1)
Putting t = -1 in (1), we get A = 1.
Putting t = -2 in (1), we get B = -1.
∴ \(\frac{1}{(1+t)(2+t)}=\frac{1}{(1+t)}-\frac{1}{(2+t)}\)
⇒ I = \(\int \frac{d t}{(1+t)(2+t)}=\int \frac{d t}{(1+t)}-\int \frac{d t}{(2+t)}\)
= log |1+t| – log |2+t| + C
= \(\log \left|\frac{1+t}{2+t}\right|+C\)
= \(\log \left|\frac{1+x^3}{2+x^3}\right|+C\)
\(\int \frac{x^2}{\left(1+x^3\right)\left(2+x^3\right)} d x\) = \(\log \left|\frac{1+x^3}{2+x^3}\right|+C\)
Example 11 Evaluate \(\int \frac{d x}{\left(e^x-1\right)}\)
Solution
Given
\(\int \frac{d x}{\left(e^x-1\right)}\)Put ex = t and exdx = dt, i.e., dx = \(\frac{1}{t}dt\), we get
I = \(\int \frac{d x}{\left(e^x-1\right)}=\int \frac{d t}{t(t-1)}\)
Let \(\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{(t-1)} .\)
Then, 1 ≡ A(t-1) + Bt …(1)
Putting t = 0 in (1), we get A = -1.
Putting t = 1 in (1), we get B = 1.
∴ \(\frac{1}{t(t-1)}=\frac{-1}{t}+\frac{1}{(t-1)}\)
Hence, I = \(\int \frac{d x}{\left(e^x-1\right)}\)
= \(\int \frac{d t}{t(t-1)}=\int \frac{-1}{t} d t+\int \frac{1}{(t-1)} d t\)
= -log |t| + log |t-1| + C
= \(\log \left|\frac{t-1}{t}\right|+C\)
= \(\log \left|\frac{e^x-1}{e^x}\right|+C\)
\(\int \frac{d x}{\left(e^x-1\right)}\) = \(\log \left|\frac{e^x-1}{e^x}\right|+C\)
Example 12 Evaluate \(\int \frac{d x}{x\left(x^n+1\right)}\)
Solution
Given
\(\int \frac{d x}{x\left(x^n+1\right)}\)Putting xn = t, we get nxn-1dx = dt.
∴ \(\frac{n x^n}{x}\) d x=d t ⇒ \(\frac{1}{x} d x=\frac{1}{n t} d t\) (note)
∴ \(\int \frac{d x}{x\left(x^n+1\right)}=\int \frac{d t}{n t(t+1)}=\frac{1}{n} \cdot \int \frac{d t}{t(t+1)}\) …(1)
Let \(\frac{1}{t(t+1)}=\frac{A}{t}+\frac{B}{(t+1)} .\)
Then, 1 ≡ A(t+1) + Bt …(2)
Putting t = 0 in (1), we get A = 1.
Putting t = -1 in (1), we get B = -1.
∴ \(\frac{1}{t(t+1)}=\left\{\frac{1}{t}-\frac{1}{(t+1)}\right\}\)
⇒ \(\int \frac{d x}{x\left(x^n+1\right)}=\frac{1}{n} \int \frac{d t}{t(t+1)}\)
= \(\frac{1}{n}\left[\int \frac{1}{t} d t-\int \frac{1}{(t+1)} d t\right]\)
= \(\frac{1}{n} \cdot\{\log |t|-\log |t+1|\}+C\)
= \(\frac{1}{n} \cdot \log \left|\frac{t}{t+1}\right|+C\)
= \(\frac{1}{n} \log \left|\frac{x^n}{x^n+1}\right|+C .\)
\(\int \frac{d x}{x\left(x^n+1\right)}\) = \(\frac{1}{n} \log \left|\frac{x^n}{x^n+1}\right|+C .\)
Applications of Partial Fraction Decomposition in Calculus .
Example 13 Evaluate \(\int \frac{d x}{x\left(x^4+1\right)}\)
Solution
Given
\(\int \frac{d x}{x\left(x^4+1\right)}\)We have
I = \(\int \frac{d x}{x\left(x^4+1\right)}=\int \frac{x^3}{x^4\left(x^4+1\right)} d x\) [multiplying num. and denom. by x3].
Putting x4 = t and 4x3dx = dt, we get
I = \(\frac{1}{4} \cdot \int \frac{d t}{t(t+1)}\)
= \(\frac{1}{4} \cdot \int\left\{\frac{1}{t}-\frac{1}{(t+1)}\right\} d t\) [by partial fraction]
= \(\frac{1}{4} \int \frac{1}{t} d t-\frac{1}{4} \int \frac{1}{(t+1)} d t\)
= \(\frac{1}{4} \log |t|-\frac{1}{4} \log |t+1|+C\)
= \(\frac{1}{4} \log \left|x^4\right|-\frac{1}{4} \log \left|x^4+1\right|+C\)
= \(\left(\frac{1}{4} \times 4\right) \log |x|-\frac{1}{4} \log \left|x^4+1\right|+C\)
= \(\log |x|-\frac{1}{4} \log \left|x^4+1\right|+C\)
\(\int \frac{d x}{x\left(x^4+1\right)}\) = \(\log |x|-\frac{1}{4} \log \left|x^4+1\right|+C\)
Example 14 Evaluate \(\int \frac{x^2}{\left(x^2+2\right)\left(x^2+3\right)} d x\)
Solution
Given
\(\int \frac{x^2}{\left(x^2+2\right)\left(x^2+3\right)} d x\)Let \(\frac{x^2}{\left(x^2+2\right)\left(x^2+3\right)}=\frac{y}{(y+2)(y+3)}\), where x2 = y.
Let \(\frac{y}{(y+2)(y+3)}=\frac{A}{(y+2)}+\frac{B}{(y+3)}\)
⇒ y ≡ A(y+3) + B(y+2) ..(1)
Putting y = -2 on both sides of (1), we get A = -2.
Putting y = -3 on both sides of (1), we get B = 3.
∴ \(\frac{y}{(y+2)(y+3)}=\frac{-2}{(y+2)}+\frac{3}{(y+3)}\)
⇒ \(\frac{x^2}{\left(x^2+2\right)\left(x^2+3\right)}=\frac{-2}{\left(x^2+2\right)}+\frac{3}{\left(x^2+3\right)}\)
⇒ \(\int \frac{x^2}{\left(x^2+2\right)\left(x^2+3\right)} d x=-2 \cdot \int \frac{d x}{\left(x^2+2\right)}+3 \cdot \int \frac{d x}{\left(x^2+3\right)}\)
= \(\frac{-2}{\sqrt{2}} \tan ^{-1}\left(\frac{x}{\sqrt{2}}\right)+\frac{3}{\sqrt{3}} \tan ^{-1}\left(\frac{x}{\sqrt{3}}\right)+C\)
= \(-\sqrt{2} \tan ^{-1}\left(\frac{x}{\sqrt{2}}\right)+\sqrt{3} \tan ^{-1}\left(\frac{x}{\sqrt{3}}\right)+C\)
\(\int \frac{x^2}{\left(x^2+2\right)\left(x^2+3\right)} d x\) = \(-\sqrt{2} \tan ^{-1}\left(\frac{x}{\sqrt{2}}\right)+\sqrt{3} \tan ^{-1}\left(\frac{x}{\sqrt{3}}\right)+C\)
Example 15 Evaluate \(\int \frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-6)} d x\)
Solution
Given
\(\int \frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-6)} d x\)Let \(\frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-6)}=1+\frac{A}{(x-4)}+\frac{B}{(x-5)}+\frac{C}{(x-6)}\). Then,
(x-1)(x-2)(x-3) ≡ (x-4)(x-5)(x-6) + A(x-5)(x-6) + B(x-4)(x-6) + C(x-4)(x-5) …(1)
Putting x = 4 on both sides of (1) we get A = 3.
Putting x = 5 on both sides of (1), we get B = -24.
Putting x = 6 on both sides of (1), we get C = 30.
∴ I = \(\int \frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-6)} d x\)
= \(\int\left\{1+\frac{3}{(x-4)}-\frac{24}{(x-5)}+\frac{30}{(x-6)}\right\} d x\)
= \(\int d x+3 \int \frac{d x}{(x-4)}-24 \int \frac{d x}{(x-5)}+30 \int \frac{d x}{(x-6)}\)
= x + 3 log |x-4| – 24 log |x-5| + 30 log |x-6| + C.
\(\int \frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-6)} d x\) = x + 3 log |x-4| – 24 log |x-5| + 30 log |x-6| + C.
Example 16 Evaluate \(\int \frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)} d x\)
Solution
Given
\(\int \frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)} d x\)We have
\(\frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)}=\frac{(t+1)(t+2)}{(t+3)(t+4)}\), where x2 = t
= \(\frac{\left(t^2+3 t+2\right)}{\left(t^2+7 t+12\right)}=1-\frac{(4 t+10)}{(t+3)(t+4)}\).
Let \(\frac{(4 t+10)}{(t+3)(t+4)}=\frac{A}{(t+3)}+\frac{B}{(t+4)}\)
⇒ (4t+10) ≡ A(t+4) + B(t+3) …(1)
Putting t =- -3 in (1), we get A = -2.
Putting t = -4 in (1), we get B = 6.
∴ \(\frac{(4 t+10)}{(t+3)(t+4)}=\frac{-2}{(t+3)}+\frac{6}{(t+4)}\) …(2)
Thus, \(\frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)}=\frac{(t+1)(t+2)}{(t+3)(t+4)}\), where x2 = t-1
= \(\frac{\left(t^2+3 t+2\right)}{\left(t^2+7 t+12\right)}=1-\frac{(4 t+10)}{(t+3)(t+4)}\)
= \(1-\left\{\frac{-2}{(t+3)}+\frac{6}{(t+4)}\right\}\) [from (2)]
= \(\left\{1+\frac{2}{(t+3)}-\frac{6}{(t+4)}\right\}=\left\{1+\frac{2}{\left(x^2+3\right)}-\frac{6}{\left(x^2+4\right)}\right\}\)
∴ \(\int \frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)} d x=\int\left\{1+\frac{2}{\left(x^2+3\right)}-\frac{6}{\left(x^2+4\right)}\right\} d x\)
= \(\int d x+2 \int \frac{d x}{\left(x^2+3\right)}-6 \int \frac{d x}{\left(x^2+4\right)}\)
= \(x+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{x}{\sqrt{3}}\right)-\frac{6}{2} \tan ^{-1}\left(\frac{x}{2}\right)+C\)
= \(x+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{x}{\sqrt{3}}\right)-3 \tan ^{-1}\left(\frac{x}{2}\right)+C\).
\(\int \frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)} d x\) = \(x+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{x}{\sqrt{3}}\right)-3 \tan ^{-1}\left(\frac{x}{2}\right)+C\).
Partial Fraction Integration Techniques for Beginners
Example 17 Evaluate \(\int \frac{(3 \sin \theta-2) \cos \theta}{\left(5-\cos ^2 \theta-4 \sin \theta\right)} d \theta\)
Solution
Given:
\(\int \frac{(3 \sin \theta-2) \cos \theta}{\left(5-\cos ^2 \theta-4 \sin \theta\right)} d \theta\)We have
I = \(\int \frac{(3 \sin \theta-2) \cos \theta}{\left\{5-\left(1-\sin ^2 \theta\right)-4 \sin \theta\right\}} d \theta\)
= \(\int \frac{(3 \sin \theta-2) \cos \theta}{\left(4+\sin ^2 \theta-4 \sin \theta\right)} d \theta\)
= \(\int \frac{(3 \sin \theta-2) \cos \theta}{(\sin \theta-2)^2} d \theta=\int \frac{(3 t-2)}{(t-2)^2} d t\), where sin θ = t.
Let \(\frac{(3 t-2)}{(t-2)^2}=\frac{A}{(t-2)}+\frac{B}{(t-2)^2}\). Then,
(3t-2) ≡ A(t-2) + B …(1)
Putting t = 2 in (1), we get B = 4.
Comparing coefficients of t on both sides of (1), we get A = 3.
Thus, A = 3 and B = 4.
∴ \(\frac{(3 t-2)}{(t-2)^2}=\frac{3}{(t-2)}+\frac{4}{(t-2)^2}\)
⇒ I = \(\int \frac{(3 t-2)}{(t-2)^2} d t=\int \frac{3}{(t-2)} d t+\int \frac{4}{(t-2)^2} d t\)
= \(3 \log |t-2|-\frac{4}{(t-2)}+C\)
= \(3 \log |\sin \theta-2|-\frac{4}{(\sin \theta-2)}+C\)
= \(3 \log (2-\sin \theta)+\frac{4}{(2-\sin \theta)}+C\) [∵ (2-sin θ) > 0].
\(\int \frac{(3 \sin \theta-2) \cos \theta}{\left(5-\cos ^2 \theta-4 \sin \theta\right)} d \theta\) = \(3 \log (2-\sin \theta)+\frac{4}{(2-\sin \theta)}+C\) [∵ (2-sin θ) > 0].
Example 18 Evaluate \(\int \frac{\left(\tan \theta+\tan ^3 \theta\right)}{\left(1+\tan ^3 \theta\right)} d \theta\)
Solution
Given
\(\int \frac{\left(\tan \theta+\tan ^3 \theta\right)}{\left(1+\tan ^3 \theta\right)} d \theta\)We have
\(\frac{\left(\tan \theta+\tan ^3 \theta\right)}{\left(1+\tan ^3 \theta\right)}=\frac{\tan \theta\left(1+\tan ^2 \theta\right)}{\left(1+\tan ^3 \theta\right)}=\frac{\tan \theta \sec ^2 \theta}{\left(1+\tan ^3 \theta\right)} .\)∴ I = \(\int \frac{\left(\tan \theta+\tan ^3 \theta\right)}{\left(1+\tan ^3 \theta\right)} d \theta\)
= \(\int \frac{\tan \theta \sec ^2 \theta}{\left(1+\tan ^3 \theta\right)} d \theta\)
= \(\int \frac{t}{\left(1+t^3\right)} d t=\int \frac{t}{(1+t)\left(1-t+t^2\right)} d t\), where tan θ = t.
Let \(\frac{t}{(1+t)\left(1-t+t^2\right)}=\frac{A}{(1+t)}+\frac{(B t+C)}{\left(1-t+t^2\right)}\). Then,
t ≡ A(1-t+t2) + (Bt+C)(1+t) …(1)
Putting t = -1 on both sides of (1), we get A = \(-\frac{1}{3}\).
Comparing coefficients of t2 on both sides of (1), we get
A + B = 0 ⇒ B = -A = \(\frac{1}{3}\).
Comparing constant terms on both sides of (1), we get
A + C = 0 ⇒ C = -A = \(\frac{1}{3}\).
∴ \(\frac{t}{(1+t)\left(1-t+t^2\right)}=\frac{-1}{3(1+t)}+\frac{\left(\frac{1}{3} t+\frac{1}{3}\right)}{\left(1-t+t^2\right)}\)
Now, I = \(\int \frac{t}{(1+t)\left(1-t+t^2\right)} d t\)
= \(-\frac{1}{3} \int \frac{d t}{(1+t)}+\frac{1}{6} \int \frac{2 t}{\left(t^2-t+1\right)} d t+\frac{1}{3} \int \frac{d t}{\left(t^2-t+1\right)}\)
= \(-\frac{1}{3} \int \frac{d t}{(1+t)}+\frac{1}{6} \int \frac{(2 t-1)+1}{\left(t^2-t+1\right)} d t+\frac{1}{3} \int \frac{d t}{\left(t^2-t+1\right)}\)
= \(-\frac{1}{3} \int \frac{d t}{(1+t)}+\frac{1}{6} \int \frac{(2 t-1)}{\left(t^2-t+1\right)} d t+\frac{1}{2} \int \frac{d t}{\left(t^2-t+1\right)}\)
= \(-\frac{1}{3} \log |1+t|+\frac{1}{6} \log \left|t^2-t+1\right|+\frac{1}{2} \int \frac{d t}{\left(t^2-t+\frac{1}{4}\right)+\frac{3}{4}}\)
= \(-\frac{1}{3} \log |1+t|+\frac{1}{6} \log \left|t^2-t+1\right|+\frac{1}{2} \int \frac{d t}{(t-1 / 2)^2+(\sqrt{3} / 2)^2}\)
= \(-\frac{1}{3} \log |1+t|+\frac{1}{6} \log \left|t^2-t+1\right|+\frac{1}{2} \cdot \frac{2}{\sqrt{3}} \tan ^{-1} \frac{\left(t-\frac{1}{2}\right)}{(\sqrt{3} / 2)}+C\)
Comparative Analysis of Different Partial Fraction Forms
= \(-\frac{1}{3} \log |1+t|+\frac{1}{6} \log \left|t^2-t+1\right|+\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 t-1}{\sqrt{3}}\right)+C\)
= \(-\frac{1}{3} \log |1+\tan \theta|+\frac{1}{6} \log \left|\tan ^2 \theta-\tan \theta+1\right|+\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan \theta-1}{\sqrt{3}}\right)+C .\)
\(\int \frac{\left(\tan \theta+\tan ^3 \theta\right)}{\left(1+\tan ^3 \theta\right)} d \theta\) = \(-\frac{1}{3} \log |1+\tan \theta|+\frac{1}{6} \log \left|\tan ^2 \theta-\tan \theta+1\right|+\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan \theta-1}{\sqrt{3}}\right)+C .\)
Example 19 Evaluate \(\int \frac{d x}{(\sin x-\sin 2 x)}\).
Solution
Given
\(\int \frac{d x}{(\sin x-\sin 2 x)}\) \(\int \frac{d x}{(\sin x-\sin 2 x)}=\int \frac{d x}{(\sin x-2 \sin x \cos x)}\)= \(\int \frac{d x}{\sin x(1-2 \cos x)}=\int \frac{\sin x}{\sin ^2 x(1-2 \cos x)} d x\)
= \(\int \frac{\sin x}{\left(1-\cos ^2 x\right)(1-2 \cos x)} d x\)
= \(-\int \frac{d t}{\left(1-t^2\right)(1-2 t)}\), where cos x = t
= \(\int \frac{d t}{(t-1)(t+1)(1-2 t)}\) …(1)
Let \(\frac{1}{(t-1)(t+1)(1-2 t)}=\frac{A}{(t-1)}+\frac{B}{(t+1)}+\frac{C}{(1-2 t)} .\)
Then, 1 ≡ A(t+1)(1-2t) + B(t-1)(1-2t) + C(t-1)(t+1) …(2)
Putting t = 1 in (2), we get A = \(-\frac{1}{2}\).
Putting t = -1 in (2), we get B = \(-\frac{1}{6}\).
Putting t = \(\frac{1}{2}\) in (2), we get C = \(-\frac{4}{3}\).
∴ I = \(-\frac{1}{2} \int \frac{d t}{(t-1)}-\frac{1}{6} \cdot \int \frac{d t}{(t+1)}-\frac{4}{3} \cdot \int \frac{d t}{(1-2 t)}\)
= \(-\frac{1}{2} \log |t-1|-\frac{1}{6} \log |t+1|+\frac{2}{3} \cdot \int \frac{-2 d t}{(1-2 t)}\)
= \(-\frac{1}{2} \log |t-1|-\frac{1}{6} \log |t+1|+\frac{2}{3} \log |1-2 t|+C\)
= \(-\frac{1}{2} \log |\cos x-1|-\frac{1}{6} \log |\cos x+1|+\frac{2}{3} \log |1-2 \cos x|+C\) .
\(\int \frac{d x}{(\sin x-\sin 2 x)}\) = \(-\frac{1}{2} \log |\cos x-1|-\frac{1}{6} \log |\cos x+1|+\frac{2}{3} \log |1-2 \cos x|+C\) .
Example 20 Evaluate \(\int \frac{(1-\cos x)}{\cos x(1+\cos x)} d x\).
Solution
Given
\(\int \frac{(1-\cos x)}{\cos x(1+\cos x)} d x\)Let \(\frac{(1-\cos x)}{\cos x(1+\cos x)}=\frac{(1-t)}{t(1+t)}=\frac{A}{t}+\frac{B}{(1+t)}\), where t = cos x.
Then, (1-t) ≡ A(1+t) + Bt.
Integration By Partial Fractions Class 12
Putting t = 0 in this identity, we get A = 1.
Putting t = -1 in this identity, we get B = -2.
∴ \(\frac{(1-t)}{t(1+t)}=\frac{1}{t}-\frac{2}{1+t}\)
or \(\frac{(1-\cos x)}{\cos x(1+\cos x)}=\frac{1}{\cos x}-\frac{2}{(1+\cos x)}\)
∴ \(\int \sec x d x-\int \sec ^2 \frac{x}{2} d x\)
= \(\log |\sec x+\tan x|-2 \tan \frac{x}{2}+C\).
\(\int \frac{(1-\cos x)}{\cos x(1+\cos x)} d x\) = \(\log |\sec x+\tan x|-2 \tan \frac{x}{2}+C\).
Example 21 Evaluate \(\int \frac{\left(x^2+1\right)}{(x+1)^2} d x\).
Solution
Given
\(\int \frac{\left(x^2+1\right)}{(x+1)^2} d x\)On dividing (x2+1) by (x2+2x+1), we get
\(\frac{\left(x^2+1\right)}{(x+1)^2}=\left\{1-\frac{2 x}{(x+1)^2}\right\} .\)Let \(\frac{2 x}{(x+1)^2}=\frac{A}{(x+1)}+\frac{B}{(x+1)^2}\)
⇒ 2x ≡ A(x+1) + B …(1)
On equating the coefficients of x, we get A = 2.
On equating constant terms, we get A + B = 0 ⇒ B = -A = -2.
∴ \(\frac{2 x}{(x+1)^2}=\frac{2}{(x+1)}-\frac{2}{(x+1)^2}\)
∴ I = \(\int\left\{1-\frac{2 x}{(x+1)^2}\right\}\)
= \(\int\left\{1-\frac{2}{(x+1)}+\frac{2}{(x+1)^2}\right\} d x\)
= \(x-2 \log |x+1|-\frac{2}{(x+1)}+C .\)
\(\int \frac{\left(x^2+1\right)}{(x+1)^2} d x\) = \(x-2 \log |x+1|-\frac{2}{(x+1)}+C .\)