Question

Evaluate the integrals:
$\int \frac{x^4}{(x^2+1)(x-1)}dx$

Answer 1

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

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

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

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

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

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

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

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

Put $x=1\Rightarrow 1=2A$ or $A=\frac{1}{2}$

Put $x=0\Rightarrow 1=A-C$ or $C=A-1=\frac{1}{2}-1=\frac{-1}{2}$

Put $x=-1\Rightarrow 1=2A+(-B+C)(-2)$

or $2A+2B-2C=1$

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

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

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

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

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

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

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

$=\frac{x^2}{2}+x+\frac{1}{2}\log (x-1)-\frac{1}{4}\log (x^2+1)-\frac{1}{2}{\tan }^{-1}x+c$ where $c$ is the constant of integration.