$\smallint \,\,{e^x}\,\,\frac{{{x^2} + \,\,1}}{{{{\left( {x + \,\,1} \right)}^2}}}\,\,dx$ is equal to

$\frac{{ - \,{e^x}}}{{x\,\, + \,\,1}}\,\, + \,\,c$

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

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

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

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

is equal to

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because the integral can be simplified and solved using integration techniques.