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Question: \[\int_{}^{}{\frac{e^{\tan^{- 1}x}}{1 + x^{2}}dx =}\]...

∫etan⁡−1x1+x2dx=\int_{}^{}{\frac{e^{\tan^{- 1}x}}{1 + x^{2}}dx =}

A

log⁡(1+x2)+c\log(1 + x^{2}) + c

B

log⁡etan⁡−1x+c\log e^{\tan^{- 1}x} + c

C

etan⁡−1x+ce^{\tan^{- 1}x} + c

D

None of these

Answer

log⁡(1+x2)+c\log(1 + x^{2}) + c

Explanation

Solution

4(sin⁡x4−cos⁡x4)+c4\left( \sin\frac{x}{4} - \cos\frac{x}{4} \right) + c

Put 4(sin⁡x4+cos⁡x4)+c4\left( \sin\frac{x}{4} + \cos\frac{x}{4} \right) + c then it reduces to

∫sin⁡x+cos⁡x1+sin⁡2x 6mudx=\int_{}^{}{\frac{\sin x + \cos x}{\sqrt{1 + \sin 2x}}\mspace{6mu} dx =}