A) \[\log 2\]
B) \[\frac{1}{2}\log 2\]
C) \[\frac{1}{2}\]
D) \[\frac{1}{\log 2}\]
Correct Answer: D
Solution :
Let \[I=\int{\frac{{{2}^{x}}}{\sqrt{1-{{4}^{x}}}}dx}\] Let \[{{2}^{x}}=t\] \[\Rightarrow \] \[{{2}^{x}}\log 2\,\,dx=dt\] \[\therefore \] \[I=\frac{1}{\log 2}\int{\frac{1}{\sqrt{1-{{t}^{{}}}}}dt}\] \[=\frac{1}{\log 2}{{\sin }^{-1}}(t)+C\] \[=\frac{1}{\log 2}{{\sin }^{-1}}({{2}^{x}})+C\] But \[I=k{{\sin }^{-1}}({{2}^{x}})+C\] \[\Rightarrow \] \[k=\frac{1}{\log 2}\]
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