Q112 / 160 · AP EAMCET 2024 (21 May Shift 2) · Mathematics

$\int \frac{x^4+1}{x^6+1} d x=$

  • A
    $\tan ^1 x-\tan ^{-1} x^3+c$
  • B
    $\tan ^{-1} x-\frac{1}{3} \tan ^{-1} x^3+c$
  • C
    $\tan ^{-1} x+\tan ^{-1} x^3+c$
  • Correct answer
    D
    $\tan ^{-1} x+\frac{1}{3} \tan ^{-1} x^3+c$

Solution

$\begin{aligned} & \text { } \frac{x^4+1}{x^6+1}=\frac{\left(x^4-x^2+1+x^2\right)}{\left(x^6+1\right)}=\frac{1}{x^2+1}+\frac{x^2}{x^6+1} \\ & \int \frac{x^4+1}{x^6+1} d x=\int \frac{d x}{1+x^2}+\int \frac{x^2 d x}{x^6+1}=\tan ^{-1} x+\frac{1}{3} \tan ^{-1} x^3+c\end{aligned}$

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

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