Question

Integrate the function w..r. to x : $\frac{4x}{2x^{2}+x-1}$

Answer 1

Solution:

First factorise the denominator,then do partial fraction and integrate

$\frac{4x}{2x^{2}+x-1}\\\\2x^{2}+x-1=2x^{2}+2x-x-1\\\\2x(x+1)-1(x+1)\\\\(2x-1)(x+1)$

now for partial fraction

$\frac{4x}{2x^{2}+x-1}=\frac{A}{(2x-1)}+\frac{B}{(x+1)}$


now find the value of constants

$A=\frac{4x}{(x+1)}\:\: put\:x=\frac{1}{2}\\\\=\frac{4}{3}$

$B=\frac{4x}{(2x-1)}\:\: put\:x=-1\\\\=\frac{4}{3}$

So,
$\frac{4x}{2x^{2}+x-1}=\frac{4}{3(2x-1)}+\frac{4}{3(x+1)}$

it is ready for integration

$\int\frac{4x}{2x^{2}+x-1}dx=\int\frac{4}{3(2x-1)}dx+\int\frac{4}{3(x+1)}dx\\\\=\frac{4 log(2x-1)}{6}+\frac{4 log(x+1)}{3}+C\\\\=\frac{2 log(2x-1)}{3}+\frac{4 log(x+1)}{3}+C$