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Q.

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

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a

$\frac{- 1}{100}$

b

$\frac{- 1}{50}$

c

$\frac{1}{100}$

d

$\frac{1}{50}$

answer is C.

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Detailed Solution

$\begin{array}{l} I = \int \frac{x^{49} \tan^{- 1} (x^{50})}{1 + x^{100}} d x \\ \tan^{- 1} x^{50} = t \\ \Rightarrow \frac{1}{1 + x^{100}} \times 50 \times x^{49} d x = d t \\ I = \frac{1}{50} \int t d t = \frac{t^{2}}{100} + c = \frac{1}{100} {(\tan^{- 1} x^{50})}^{2} + c \end{array}$

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