Question

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

Answer 1

$I=\int \frac{x^9}{(x-1)(x^2+1)}dx$

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

$=\int \underset{I_1}{\underset{}{\frac{(x^2{)}^2-1}{(x-1)(x^2+1)}}}dx+\underset{I_2}{\underset{}{\int \frac{1}{(x-1)(x^2+1)}}}dx$

$I_1=\int \frac{(x^2-1)(x^2+1)}{(x-1)(x^2+1)}dx=\int (x+1)dx=\frac{x^2}{2}+x+C_1$

$I_2=\frac{1}{(x-1)(x^2+1)}=\frac{A}{(x-1)}+\frac{Bx+C}{x^2+1}$

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

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

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

On comparison, we get

$A+B=0$ or $A=-B$ $\to \left(1\right)$

$C=0$ $\to \left(2\right)$

$A-B-C=1\to \left(3\right)$

$-B-B-0=1$ or $-2B=1$ $\therefore B=\frac{-1}{2}$

$\therefore A=-B=\frac{+1}{2}$ and $C=0$

Then

$I_2=\int [\frac{\frac{1}{2}}{x-1}+\frac{\frac{-1}{2}x}{x^2+1}]dx$

$=\frac{1}{2}log|x-1|-\frac{1}{4}log|x^2+1|+C_2$

$\therefore I=\frac{x^2}{2}+x+\frac{1}{2}log|x-1|-\frac{1}{4}log|x^2+1|+C$.