Question

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

• A. $\ell n2$
• B. $\frac{1}{2}\ell n2$
• C. $\frac{1}{2}$
• D. $\frac{1}{\ell n2}$

Answer 1

The correct option is D $\frac{1}{\ell n2}$

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

put $t=2^x⟹dt=2^x\ln 2dx⟹dx=\frac{dt}{t\ln 2}$

$\int \frac{1}{\sqrt{1-t^2}}\times \frac{dt}{\ln 2}=\frac{1}{\ln 2}{\text{sin}}^{-1}(2{)}^x+C⟹K=\frac{1}{\ln 2}$