Q110 / 160 · TS EAMCET 2023 (14 May Shift 1) · Mathematics

If $\int \frac{x^{49} \tan ^{-1}\left(x^{50}\right)}{\left(1+x^{100}\right)} d x=k\left(\tan ^{-1}\left(x^{50}\right)\right)^2+c$, then $k=$

  • A
    $\frac{-1}{100}$
  • B
    $\frac{1}{50}$
  • C
    $\frac{-1}{50}$
  • Correct answer
    D
    $\frac{1}{100}$

Solution

$\int \frac{x^{49} \tan ^{-1} x^{50}}{1+x^{100}} d x$ Let $\tan ^{-1} x^{50}=\mathrm{t}$ $$ \begin{aligned} & \frac{50 x^{49}}{1+\left(x^{50}\right)^2} d x=d t \\ & \frac{1}{50} \int t d t=\frac{1}{50} \frac{t^2}{2}+\mathrm{C}=\frac{1}{100}\left(\tan ^{-1} x^{50}\right)^2+\mathrm{C} \\ & \therefore k=\frac{1}{100} \end{aligned} $$

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

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