Question

Integrate $\int \frac{x^{3}dx}{(x-1)(x^2+1)}$

Answer 1

Solving by partial fractions,

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

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

Put $x=1\Rightarrow 1=A(1+1)+(B+C)(1-1)$

$\Rightarrow 2A=1$

$\Rightarrow A=\frac{1}{2}$

Put $x=0\Rightarrow 0=A(0+1)+(0+C)(0-1)$

$\Rightarrow A-C=0$

$\Rightarrow C=A=\frac{1}{2}$ since $A=\frac{1}{2}$

Put $x=-1\Rightarrow -1=A(1+1)+(-B+C)(-1-1)$

$\Rightarrow -1=2A+2B-2C$

$\Rightarrow -1=2\times \frac{1}{2}+2B-2\times \frac{1}{2}$

$\Rightarrow -1=1+2B-1$

$\Rightarrow 2B=-1$

$\Rightarrow B=\frac{-1}{2}$

$\therefore A=\frac{1}{2},B=\frac{-1}{2},C=\frac{1}{2}$

Now ,$\frac{x^3}{(x-1)(x^2+1)}$

$=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}$

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

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

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

Integrating w.r.t $x$ we get

$=\frac{1}{2}\int \frac{dx}{x-1}+\frac{1}{4}\int \frac{2xdx}{x^2+1}+\frac{1}{2}\int \frac{1}{x^2+1}$

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