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

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Correct option is A

I=∫e2log⁡x1+x6−1dxI=∫elog⁡x21+x6dx=13∫3x21+x6dx Let x3=t→3x2dx=dtI=13∫11+t2dt=13tan−1⁡(t)+CI=13tan−1⁡x3+C

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∫e2logx1+x6−1dx=−−−

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