Question

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

Answer 1

$\begin{array}{c}\int \frac{x^4}{\left(x-1\right)\left(x^2+1\right)}dx\\ \frac{x^4}{\left(x-1\right)\left(x^2+1\right)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}\\ \Rightarrow x^4=A\left(x^2+1\right)\left(Bx+C\right)\left(x-1\right)\\ puttingx=1onbothsides\\ 1=2A\\ A=\frac{1}{2}\\ puttingx=0onbothsides\\ 0=A-c\\ c=A=\frac{1}{2}\\ puttingx=-1onbothsides\\ 1=2A+\left(-B+c\right)\left(-2\right)\\ \Rightarrow 1=1+2B-2c\\ \Rightarrow 2c=2B\\ B=c=\frac{1}{2}\\ so,\\ \int \frac{x^4}{\left(x-1\right)\left(x^2+1\right)}=\frac{1}{2}\int \frac{dx}{x-1}+\frac{1}{2}\int \frac{x+1}{x^2+1}dx\\ =\frac{1}{2}\int \frac{dx}{x-1}+\frac{1}{4}\int \frac{2x}{x^2+1}dx+\frac{1}{2}\int \frac{dx}{x^2+1}\\ =\frac{1}{2}\log \left|x-1\right|+\frac{1}{4}\log \left(x^2+1\right)+\frac{1}{2}{\tan }^{-1}x+c\end{array}$