Question

If $\int \frac{2^x}{\sqrt{1-4^x}}dx=k{\sin }^{-1}\left(2^x\right)+C$, then $k$ is equal to

• A. $\log 2$
• B. $\frac{1}{2}\log 2$
• C. $\frac{1}{2}$
• D. $\frac{1}{\log 2}$

Answer 1

The correct option is D $\frac{1}{\log 2}$

Let $I=\int \frac{2^x}{\sqrt{1-4^x}}dx$

Let $2^x=t$

$\Rightarrow 2^x\log 2dx=dt$

Thus $I=\frac{1}{\log 2}\int \frac{1}{\sqrt{1-t^2}}dt$

$=\frac{1}{\log 2}{\sin }^{-1}\left(t\right)+C$

$=\frac{1}{\log 2}{\sin }^{-1}\left(2^x\right)+C$

But $I=k{\sin }^{-1}\left(2^x\right)+C$,

$\Rightarrow k=\frac{1}{\log 2}$.