Mathematics - Indefinite Integration Question with Solution | TestHub

$I=\int \left(\frac{x^3-1}{x^3+x}\right)dx=\int \left(1-\frac{x+1}{x^3+x}\right)dx$ $\begin{array}{rl}& \Rightarrow I=\int 1\cdot dx-\int \frac{(x+1)}{x^3+x}dx\\ & =x-\int \frac{(x+1)}{x\left(x^2+1\right)}dx\end{array}$ $\begin{array}{rl}& \text{ Let }\frac{x+1}{x\left(x^2+1\right)}=\frac{A}{x}+\frac{Bx+C}{x^2+1}\\ & \Rightarrow (x+1)=A\left(x^2+1\right)+(Bx+C)x\\ & \Rightarrow (x+1)=(A+B)x^2+Cx+A\end{array}$ On comparing coefficient, we get $\begin{array}{rl}& A+B=0,C=1,A=1\\ & \Rightarrow B=-1\\ & \therefore I=x-\int \frac{1}{x}dx-\int \frac{(1-x)}{x^2+1}dx\\ & \Rightarrow I=x-\log |x|-\int \frac{1}{x^2+1}dx+\frac{1}{2}\int \frac{2x}{x^2+1}dx\\ & \Rightarrow I=x-\log |x|-{\tan }^{-1}x+\frac{1}{2}\log \left(x^2+1\right)+c\end{array}$