Q58 / 100 · BITSAT 2024 (Memory Based Paper 2) · Mathematics
$\int \frac{x^3-1}{x^3+x} d x=$
- A$x+\log |x|+\frac{1}{2} \log \left(x^2+1\right)+\sin ^{-1}(x)+c$
- B$x-\log |x|+\frac{1}{2} \log \left(x^2+1\right)-\sin ^{-1}(x)+c$
- C$x+\log |x|-\frac{1}{2} \log \left(x^2+1\right)+\tan ^{-1}(x)+c$
- Correct answerD$x-\log |x|+\frac{1}{2} \log \left(x^2+1\right)-\tan ^{-1}(x)+c$
Solution
$I=\int\left(\frac{x^3-1}{x^3+x}\right) d x=\int\left(1-\frac{x+1}{x^3+x}\right) d x$
$\begin{aligned}
& \Rightarrow I=\int 1 \cdot d x-\int \frac{(x+1)}{x^3+x} d x \\
& =x-\int \frac{(x+1)}{x\left(x^2+1\right)} d x
\end{aligned}$
$\begin{aligned}
& \text { Let } \frac{x+1}{x\left(x^2+1\right)}=\frac{A}{x}+\frac{B x+C}{x^2+1} \\
& \Rightarrow(x+1)=A\left(x^2+1\right)+(B x+C) x \\
& \Rightarrow(x+1)=(A+B) x^2+C x+A
\end{aligned}$
On comparing coefficient, we get
$\begin{aligned}
& A+B=0, C=1, A=1 \\
& \Rightarrow B=-1 \\
& \therefore I=x-\int \frac{1}{x} d x-\int \frac{(1-x)}{x^2+1} d x \\
& \Rightarrow I=x-\log |x|-\int \frac{1}{x^2+1} d x+\frac{1}{2} \int \frac{2 x}{x^2+1} d x \\
& \Rightarrow I=x-\log |x|-\tan ^{-1} x+\frac{1}{2} \log \left(x^2+1\right)+c
\end{aligned}$
Hence, the correct answer is option: 'D'.