The anti-derivative of $\frac{{{2^x}}}{{\sqrt {1 - {4^x}} }}$ w.r.t.$x$ is

${\cos ^{ - 1}}\left( {{2^x}} \right){\log _2}e + k$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

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Application +4 | -1

The anti-derivative of

\frac{{{2^x}}}{{\sqrt {1 - {4^x}} }}

w.r.t.

x

View answer and explanation

Correct option: A

{\log _2}e \cdot {\sin ^{ - 1}}\left( {{2^x}} \right) + k

Option C (

{\cos ^2}x\left( {\frac{1}{2} - \log \cos x} \right) + k

) is correct because after integrating by parts and simplifying, the result matches this option.

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