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

The Correct Option is 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 $t={{e}^{2x}}$ $dt=2{{e}^{2x}}\,dx$
$\Rightarrow$ $\int{\frac{dt}{2(1+{{t}^{2}})}}=\frac{1}{2}{{\tan }^{-1}}t+C$
$\Rightarrow$ $=\frac{1}{2}{{\tan }^{-1}}({{e}^{2x}})+C$