$\smallint \frac{{{e^{2x}} - \,\,2{e^x}}}{{{e^{2x}} + 1}}\,\,dx$ is equal to

$\log \,\,\left( {{e^{2x}} + 1} \right) - \,\,{\tan ^{ - 1}}\,\,\left( {{e^x}} \right) + C$

$\frac{1}{2}\,\,\log \,\,\left( {{e^{2x}} + \,\,1} \right) - {\tan ^{ - 1}}\,\,\left( {{e^x}} \right) + C$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

\int \frac{{{e^{2x}} - \,\,2{e^x}}}{{{e^{2x}} + 1}}\,\,dx

is equal to

View answer and explanation

Correct option: C

\frac{1}{2}\,\,\log \,\,\left( {{e^{2x}} + 1} \right) - 2\,\,{\tan ^{ - 1}}\,\,\left( {{e^x}} \right) + C

Option C (

\frac{{1}}{{2}}\,\log \,\left( {{e^{2x}} + 1} \right) - 2\,{\tan ^{ - 1}}\,\left( {{e^x}} \right) + C

) is correct because it is the result of direct integration after separating the original integral into two simpler integrals.