Q110 / 160 · AP EAMCET 2024 (21 May Shift 2) · Mathematics

$\int \frac{x^5}{x^2+1} d x=$

  • A
    $\frac{x^4}{4}+\frac{x^3}{3}-\tan ^{-1} x+c$
  • Correct answer
    B
    $\frac{x^4}{4}-\frac{x^2}{2}+\frac{1}{2} \log \left(x^2+1\right)+c$
  • C
    $\frac{x^4}{4}+\frac{x^3}{3}+\tan ^{-1} x+c$
  • D
    $\frac{x^4}{4}+\frac{x^2}{2}-\frac{1}{2} \log \left(x^2+1\right)+c$

Solution

$\begin{aligned} & \text { } I=\int \frac{x^5}{x^2+1} d x=\int \frac{\left(x^4-1+1\right) x d x}{\left(x^2+1\right)} \\ & =\int\left(x^2-1+\frac{1}{x^2+1}\right) d x=\frac{x^4}{4}-\frac{x^2}{2}+\frac{1}{2} \log \left(x^2+1\right)+c\end{aligned}$

Hence, the correct answer is option: 'B'.

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