Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Q10 Integrate the rational functions $\frac{2x -3 }{(x^2 -1 )( 2x-3)}$

Answers (1)

best_answer

Given function $\frac{2x -3 }{(x^2 -1 )( 2x-3)}$

can be rewritten as $\frac{2x -3 }{(x^2 -1 )( 2x-3)} = \frac{2x-3}{(x+1)(x-1)(2x+3)}$

The partial function of this function:

$\frac{2x -3 }{(x^2 -1 )( 2x-3)} = \frac{A}{(x+1)} +\frac{B}{(x-1)}+\frac{C}{(2x+3)}$

$\Rightarrow (2x-3) =A(x-1)(2x+3)+B(x+1)(2x+3)+C(x+1)(x-1)$ $\Rightarrow (2x-3) =A(2x^2+x-3)+B(2x^2+5x+3)+C(x^2-1)$ $\Rightarrow (2x-3) =(2A+2B+C)x^2+(A+5B)x+(-3A+3B-C)$

Equating the coefficients of $x^2\ and\ x$ , we get

$B=-\frac{1}{10},\ A =\frac{5}{2},\ and\ C= -\frac{24}{5}$

Therefore,

$\frac{2x -3 }{(x^2 -1 )( 2x-3)} = \frac{5}{2(x+1)} -\frac{1}{10(x-1)}-\frac{24}{5(2x+3)}$

$\implies \int \frac{2x-3}{(x^2-1)(2x+3)}dx = \frac{5}{2}\int \frac{1}{(x+1)}dx -\frac{1}{10}\int \frac{1}{x-1}dx -\frac{24}{5}\int \frac{1}{(2x+3)}dx$ $= \frac{5}{2}\log|x+1| -\frac{1}{10}\log|x-1| -\frac{24}{10}\log|2x+3|$

$= \frac{5}{2}\log|x+1| -\frac{1}{10}\log|x-1| -\frac{12}{5}\log|2x+3|+C$

$= -\frac{1}{2}\log|x-1| +4\left ( \frac{-1}{x-1} \right ) +\frac{1}{2}\log|x+1| +C$

$=\frac{1}{2}\log|\frac{x+1}{x-1}| - \frac{4}{(x-1)} + +C$

$= \frac{2}{9}\log|x-1|+\frac{1}{3}\left ( \frac{-1}{x-1} \right )-\frac{2}{9}\log|x+2|+C$

$\frac{2}{9}\log\left | \frac{x-1}{x+2} \right | -\frac{1}{3(x-1)}+C$

Posted by

Divya Prakash Singh

View full answer

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE explains 'roll number not found' error on portal amid student concerns

anam-khan

Congress attacks Dharmendra Pradhan over CBSE row, alleges failures in exam and revaluation process

anam-khan

CBSE retains Coempt for Class 12 revaluation amid OSM controversy, shifts data to its own servers: Reports

Latest Question