$\int \frac{x^3-1}{x^3+x} d x=$

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Aicharya · Accepted Answer

$\begin{aligned} & I=\int\left(\frac{x^3-1}{x^3+x}\right) d x=\int\left(1-\frac{x+1}{x^3+x}\right) d x \ & \Rightarrow \int 1 \cdot d x-\int \frac{(x+1)}{x^3+x} d x=x-\int \frac{(x+1)}{x\left(x^2+1\right)} d x \end{aligned}$ Now, $\frac{x+1}{x\left(x^2+1\right)}=\frac{A}{x}+\frac{B x+C}{x^2+1}$ $\Rightarrow \quad(x+1)=A\left(x^2+1\right)+(B x+C) x$ [using partial fractions] $\Rightarrow \quad(x+1)=(A+B) x^2+C x+A$ On comparing $A+B=0, C=1, A=1$ $\begin{array}{lc} \Rightarrow & B=-1 \ \therefore & I=x-\int \frac{1}{x} d x-\int \frac{(1-x)}{x^2+1} d x \ \Rightarrow & I=x-\log |x|-\int \frac{1}{x^2+1} d x+\frac{1}{2} \int \frac{2 x}{x^2+1} d x \ \Rightarrow & I=x-\log |x|-\tan ^{-1} x+\frac{1}{2} \log \left(x^2+1\right)+C \end{array}$

\(\int \frac{x^3-1}{x^3+x} d x=\)

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