A) \[2\,{{\tan }^{-1}}\,({{e}^{2x}})+C\]
B) \[{{\tan }^{-1}}\,({{e}^{2x}})+C\]
C) \[\frac{1}{2}{{\tan }^{-1}}\,({{e}^{2x}})+C\]
D) \[\frac{-1}{{{({{e}^{2x}}+{{e}^{-2x}})}^{2}}}+C\]
Correct Answer: C
Solution :
\[\int{\frac{dx}{{{e}^{2x}}+{{e}^{-2x}}}}\Rightarrow \,\int{\frac{{{e}^{2x}}}{{{e}^{4x}}+1}}dx\] \[=\int{\frac{{{e}^{2x}}}{1+{{({{e}^{2x}})}^{2}}}}dx,\] put \[t={{e}^{2x}}\] \[dt=2{{e}^{2x}}\,dx\] \[\Rightarrow \] \[\int{\frac{dt}{2(1+{{t}^{2}})}}=\frac{1}{2}{{\tan }^{-1}}t+C\] \[\Rightarrow \] \[=\frac{1}{2}{{\tan }^{-1}}({{e}^{2x}})+C\]
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