Question

integrate the rational function 2x-3/(x2 -1) (2x+3) please answer if confused look at class 12maths exercise 7.5 question number 10​

Answer 1

Given :-

$\sf \dfrac{2x-3}{(x^{2}-1(2x+3)} =\dfrac{2x-3}{(x+1)(x-1)(2x+3)}$

Solution :-

We know that,

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

Also as,

$\sf (2x-3)=A(x-1)(2x+3)+ B(x+1)(2x+3)+C(x+1)(x-1)$

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

Now, we get

$\sf (2x-3)=(2A+2B+2C)x^{2}+(A+5B)x+(-A+3B-C) \quad ....(1)$

By equating coefficients of x²,

$\sf B = \dfrac{-1}{10}$

$\sf A=\dfrac{5}{2}$

$\sf C=\dfrac{-24}{5}$

Now we get,

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

By integrating,

$\sf \int \dfrac{2x-3}{(x^{2}-1)(2x+3)} dx=\dfrac{5}{2} \int \dfrac{1}{(x+1)} dx-\dfrac{1}{10} \int \dfrac{1}{x-1} dx-\dfrac{24}{5} \int \dfrac{1}{(2x+3)} dx$

Now, here

$\sf =\dfrac{5}{2} log \right{|} x+1 \left{|}-\dfrac{1}{10} log \right{|} x-1 \left{|}-\dfrac{24}{5 \times 2} log \right{|} 2x+3 \left{|}$

$\sf =\dfrac{5}{2} log \right{|} x+1 \left{|} -\dfrac{1}{10} log \right{|}x-1 \left{|}-\dfrac{12}{5} log \right{|} 2x+3 \left{|} +C$