The value of the integral $\smallint \,\,{e^x}\,\,{\left( {\frac{{1 - x}}{{1 + x}}} \right)^2}\,dx$ is

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

${e^x}\,\,\left( {\frac{{1 + x}}{{1 + {x^2}}}} \right)\,\, + \,\,c$

${e^x}\,\,\left( {1 - x} \right)\,\, + \,\,c$

Back to INDEFINITE INTEGRATION

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Medium

1 of 1

Analyzing +4 | -1

The value of the integral

\int \,\,{e^x}\,\,{\left( {\frac{{1 - x}}{{1 + x}}} \right)^2}\,dx

View answer and explanation

Correct option: C

\frac{{{e^x}}}{{1 + {x^2}}}\,\, + \,\,c

Option C is correct because the integral can be solved by rewriting the integrand and using integration by parts, leading to the result e^x / (1 + x²) + c.

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

More Questions from INDEFINITE INTEGRATION