Question

The value of the integral $ \int{\frac{1}{{{e}^{2x}}+{{e}^{-2x}}}}\,\,dx $ is equal to

• 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 $

Answer 1

Answer: (C)

$ \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 $