Question

lf $\int \frac{2^x}{\sqrt{1-4^x}}$ dx $=k{\sin }^{-1}{2}^x+c$, then $k=$

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

Answer 1

Answer: (B)

$\frac{1}{\log 2}$

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

Put $2^x=t$

$2^x\log 2dx=dt$

$2^{x}dx=\frac{dt}{\log 2}$

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

$=\frac{1}{\log 2}{\sin }^{-1}\left(t\right)+c$ ........ $\int \frac{1}{\sqrt{a^2-x^2}}dx={\sin }^{-1}\left(\frac{x}{a}\right)+c$

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

So, $k=\frac{1}{\log 2}$