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

${\log _e}\,\,\frac{1}{{{x^2} + 1}}\,\, + \,\,c$

${\log _e}\,\,\left\{ {x\,\,\left( {{x^2} + \,\,1} \right)} \right\} + c$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

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

is equal to

View answer and explanation

Correct option: B

{\log _e}\,\,x\,\, + \,\,2\,\,{\tan ^{ - 1}}\,\,x\,\, + \,\,c

Option B (

{\log _e}\,\,x\,\, + \,\,2\,\,{\tan ^{ - 1}}\,\,x\,\, + \,\,c

) is correct because the integral simplifies to the sum of the integral of 1/x and 2/(x²⁺1).