Question

The antiderivative of $\frac{2^x}{\sqrt{1-4^x}}$ w.r.t. $x$ is

• A. ${\log }_{2}e.{\sin }^{-1}\left(2^x\right)+k$
• B. ${\sin }^{-1}\left(2^x\right)+k$
• C. ${\cos }^{-1}\left(2^x\right).\frac{1}{{\log }_{e}2}+k$
• D. none of these

Answer 1

Answer: (A)

${\log }_{2}e.{\sin }^{-1}\left(2^x\right)+k$

$\int \frac{2^x}{\sqrt{1-4^x}}dx$
Put $2^x=t$
$2^x{\log }_{e}2dx=dt$
$=\int \frac{1}{{\log }_{e}2\sqrt{1-t^2}}dx$
$={\log }_{2}e{\sin }^{-1}t+k$
$={\log }_{2}e{\sin }^{-1}\left(2^x\right)+k$