Question

$ \int\frac{2^{x}}{\sqrt{1-4^{x}}}dx $ is equal to

• A. $ (log\,2)\sin^{-1} 2^x + C $
• B. $ \frac{1}{2}\sin^{-1} 2^x + C $
• C. $ \frac{1}{log 2}\sin^{-1} 2^x + C $
• D. $ 2log 2\sin^{-1} 2^x+C $

Answer 1

Answer: (C)

Let $I=\int \frac{2^{x}}{\sqrt{1-4^{x}}} d x$
$=\int \frac{2^{x}}{\sqrt{1-\left(2^{x}\right)^{2}}} d x$
Put $2^{x}=t$
$\Rightarrow 2^{x} \log 2 d x=d t$
$\therefore I=\int \frac{\frac{1}{\log 2}}{\sqrt{1-t^{2}}} d t$
$=\frac{1}{\log 2} \cdot \sin ^{-1} t+C$
$=\frac{1}{\log 2} \sin ^{-1} 2^{x}+C$