Question

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

Answer 1

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

$$\begin{gathered} \frac{x^2}{(x-1)(x^2+1)}=\frac{A}{(x-1)}+\frac{Bx+C}{x^2+1} \\ =\frac{A\left(x^2+1\right)+\left(Bx+C\right)\left(x-1\right)}{(x-1)(x^2+1)} \\ \Rightarrow \frac{x^2}{(x-1)(x^2+1)}=\frac{(A+B)x^2+(C-B)x+\left(A-C\right)}{(x-1)(x^2+1)} \end{gathered}$$

Comparing coefficients, we get

$$\begin{gathered} A+B=1;C-B=0\mathrm{and}A-C=0 \\ \mathrm{Solving}\mathrm{these}\mathrm{equations},\mathrm{we}\mathrm{get} \\ A=B=C=\frac{1}{2} \\ \therefore I=\frac{1}{2}\int \frac{1}{(x-1)}dx+\frac{1}{2}\int \frac{x}{x^2+1}dx+\frac{1}{2}\int \frac{1}{x^2+1}dx \\ =\frac{1}{2}\ln \left|x-1\right|+\frac{1}{4}\ln \left|x^2+1\right|+\frac{1}{2}{\tan }^{-1}\left(x\right)+c \end{gathered}$$