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

$\frac{{\sqrt {{x^2}\,\, + \,\,2x\,\, + \,\,2} }}{{{{\left( {x\,\, + \,\,1} \right)}^2}}}\,\, + \,\,C$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

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

is equal to

View answer and explanation

Correct option: C

\frac{{ - \,\sqrt {{x^2}\,\, + \,\,2x\,\, + \,\,2} }}{{x\,\, + \,\,1}}\,\, + \,\,C

Option C (

\frac{{ - \,\sqrt {{{{x^2}\,\, + \,\,2x\,\, + \,\,2}}} }}{{{{x\,\, + \,\,1}}}}\,\, + \,\,C

) is correct because after substitution and integration, the result matches this option.