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Question: \(\int_{}^{}{\frac{a^{x}}{\sqrt{1 - a^{2x}}}dx =}\) is equal to....

∫ax1−a2xdx=\int_{}^{}{\frac{a^{x}}{\sqrt{1 - a^{2x}}}dx =} is equal to.

A

1log⁡asin⁡−1ax+c\frac{1}{\log a}\sin^{- 1}a^{x} + c

B

sin⁡−1ax+c\sin^{- 1}a^{x} + c

C

1log⁡acos⁡−1ax+c\frac{1}{\log a}\cos^{- 1}a^{x} + c

D

None of these

Answer

sin⁡−1ax+c\sin^{- 1}a^{x} + c

Explanation

Solution

2x+1−log⁡(x+1)+c\frac{2}{x + 1} - \log(x + 1) + c

∫cosecθ−cot⁡θcosecθ+cot⁡θ 6mudθ=\int_{}^{}\frac{\text{cosec}\theta - \cot\theta}{\text{cosec}\theta + \cot\theta}\mspace{6mu} d\theta = 2cosecθ−2cot⁡θ−θ+c2\text{cosec}\theta - 2\cot\theta - \theta + c.