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Q.

$\int e^{2 \log x} {(1 + x^{6})}^{- 1} d x = - - -$

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a

$\frac{1}{6} \tan^{- 1} (x^{6}) + C$

b

$\frac{1}{3} \tan^{- 1} (x^{3}) + C$

c

None

d

$\frac{1}{2} \tan^{- 1} (x^{2}) + C$

answer is A.

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Detailed Solution

$\begin{array}{l} I = \int e^{2 \log x} {(1 + x^{6})}^{- 1} d x \\ I = \int \frac{e^{\log (x^{2})}}{1 + x^{6}} d x = \frac{1}{3} \int \frac{3 x^{2}}{1 + x^{6}} d x \\ Let x^{3} = t \to 3 x^{2} d x = d t \\ I = \frac{1}{3} \int \frac{1}{1 + t^{2}} d t = \frac{1}{3} \tan^{- 1} (t) + C \\ I = \frac{1}{3} \tan^{- 1} (x^{3}) + C \end{array}$

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