A) \[(log2)si{{n}^{-1}}{{2}^{x}}+C\]
B) \[\frac{1}{2}{{\sin }^{-1}}{{2}^{x}}+C\]
C) \[\frac{1}{\log 2}{{\sin }^{-1}}{{2}^{x}}+C\]
D) \[2\log 2{{\sin }^{-1}}{{2}^{x}}+C\]
Correct Answer: C
Solution :
[c] :Let \[I=\int_{{}}^{{}}{\frac{{{2}^{x}}}{\sqrt{1-{{4}^{x}}}}}dx\] Put \[{{2}^{x}}=t\Rightarrow {{2}^{x}}\log 2\,dx=dt\] \[\therefore \]\[I=\frac{1}{\log 2}\int_{{}}^{{}}{\frac{dt}{\sqrt{1-{{t}^{2}}}}=\frac{1}{\log 2}{{\sin }^{-1}}\frac{t}{1}+C}\] \[=\frac{1}{\log 2}{{\sin }^{-1}}{{2}^{x}}+C\]You need to login to perform this action.
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