# Evaluate the Following Integral ; ∫ X ( X 2 + 1 ) ( X − 1 ) D X - Mathematics

Sum

Evaluate the following integral :-

$\int\frac{x}{\left( x^2 + 1 \right)\left( x - 1 \right)}dx$

#### Solution

$\text{Let }I = \int\frac{x}{\left( x^2 + 1 \right)\left( x - 1 \right)}dx$

We express

$\frac{x}{\left( x^2 + 1 \right)\left( x - 1 \right)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 1}$

$\Rightarrow x = A\left( x^2 + 1 \right) + \left( Bx + C \right)\left( x - 1 \right)$

Equating the coefficients of x^2 , x and constants, we get

$0 = A + B\text{ and }1 = - B + C\text{ and }0 = A - C$

$\text{or }A = \frac{1}{2}\text{ and }B = - \frac{1}{2}\text{ and }C = \frac{1}{2}$

$\therefore I = \int\left( \frac{\frac{1}{2}}{x - 1} + \frac{- \frac{1}{2}x + \frac{1}{2}}{x^2 + 1} \right)dx$

$= \frac{1}{2}\int\frac{1}{x - 1}dx - \frac{1}{2}\int\frac{x}{x^2 + 1} dx + \frac{1}{2}\int\frac{1}{x^2 + 1} dx$

$= \frac{1}{2} I_1 - \frac{1}{2} I_2 + \frac{1}{2} I_3 ............(1)$

$\text{Now, }I_1 = \int\frac{1}{x - 1}dx$

Let x - 1 = u

On differentiating both sides, we get

$dx = du$

$\therefore I_1 = \int\frac{1}{u}du$

$= \log\left| u \right| + c_1$

$= \log\left| x - 1 \right| + c_1 ..............(2)$
$\text{And, }I_2 = \int\frac{x}{x^2 + 1} dx$
$\text{Let }\left( x^2 + 1 \right) = u$

On differentiating both sides, we get
$2x\ dx = du$

$\therefore I_2 = \frac{1}{2}\int\frac{1}{u}du$

$= \frac{1}{2}\log\left| u \right| + c_2$

$= \frac{1}{2}\log\left| x^2 + 1 \right| + c_2 .............(3)$

$\text{And, }I_3 = \int\frac{1}{x^2 + 1} dx$

$= \tan^{- 1} x + c_3 ..............(4)$

From (1), (2), (3) and (4), we get

$\therefore I = \frac{1}{2}\left( \log\left| x - 1 \right| + c_1 \right) - \frac{1}{2}\left( \frac{1}{2}\log\left| x^2 + 1 \right| + c_2 \right) + \frac{1}{2}\left( \tan^{- 1} x + c_3 \right)$

$= \frac{1}{2}\log\left| x - 1 \right| - \frac{1}{4}\log\left| x^2 + 1 \right| + \frac{1}{2} \tan^{- 1} x + c$

$\text{Hence, }\int\frac{x}{\left( x^2 + 1 \right)\left( x - 1 \right)}dx = \frac{1}{2}\log\left| x - 1 \right| - \frac{1}{4}\log\left| x^2 + 1 \right| + \frac{1}{2} \tan^{- 1} x + c$

Concept: Evaluation of Simple Integrals of the Following Types and Problems
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Q 16 | Page 176