Question

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

• A. $(log2){\sin }^{-1}{2}^x+C$
• B. $\frac{1}{2}{\sin }^{-1}{2}^x+C$
• C. $\frac{1}{log2}{\sin }^{-1}{2}^x+C$
• D. $2log2{\sin }^{-1}{2}^x+C$

Answer 1

Answer: (C)

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