Question

Evaluate $\int \frac{x}{x^4-1}dx$

Answer 1

Given :

$\int \left(\frac{x}{x^4-1}\right)$

Sol.:

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

we can solve it as:-

$\frac{Ax+B}{x^2+1}+\frac{C}{x+1}+\frac{D}{x-1}+\frac{(Ax+B)(x^2-1)+C(x^2+1)(x-1)+D(x^2+1)(x+1)}{(x-1)(x+1)(x^2+1)}$

$=\frac{A{x}^3+B{x}^2-Ax-B+C{x}^3-C{x}^2+Cx-C+D{x}^3+D{x}^2+Dx+D}{x^4-1}$

$=\frac{x^3(A+C+D)+x^2(B-C+D)+x(-A+C+D)+(-B-C+D)}{(x^4-1)}$

From comparing the equation with \

$\frac{x}{x^4-1};$

we get

$A+C+D=0$

$B-C+D=0$

$-A+C+D=1$

&$(-B-C+D)=0$

$\Rightarrow A=\frac{-1}{2},B=0,C=D=\frac{1}{4}$

$\int \left(\frac{x}{x^4-1}\right)dx=\int \left(\right(\frac{1}{2}\left)\right(\frac{x}{x^2+1}\left)\right)dx+\int \left(\right(\frac{1}{4}\left)\right(\frac{1}{x+1}+\frac{1}{x-1}\left)\right)dx$

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

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

Hence, correct answer is

$\frac{1}{4}\left(ln\right(\left|\right(x^4-1\left)\right|\left)\right)+C$