Question

$\int \frac{2^x}{\left(\sqrt{1-4^x}\right)}dx=$

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

Answer 1

The correct option is C

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

Explanation for Correct answer:

Finding the value for the given function:

$\int \frac{2^x}{\left(\sqrt{1-4^x}\right)}dx=\int \frac{2^x}{\left(\sqrt{1-{\left(2^x\right)}^2}\right)}dx$

Solving this by using the substitution method

$$\begin{gathered} t=2^x \\ dt=2^x\log 2dx\left[\because \frac{d}{dx}{2}^x=2^x{\log }_{e}2\right] \\ {2}^{x}dx=\frac{1}{\log 2}dt \end{gathered}$$

Substituting the above values in the equation we get,

$$\begin{gathered} \int \frac{2^x}{\left(\sqrt{1-{\left(2^x\right)}^2}\right)}dx=\left(\frac{1}{\log 2}\right)\int \frac{dt}{\left(\sqrt{1-{\left(t\right)}^2}\right)} \\ =\left(\frac{1}{\log 2}\right){\sin }^{-1}t+C\left[\because \int \frac{1}{\sqrt{1-x^2}}={\sin }^{-1}x\right] \\ =\left(\frac{1}{\log 2}\right){\sin }^{-1}{2}^x+C \end{gathered}$$

Hence, option (C) is the correct answer.