$\smallint \frac{{x + 2}}{{\left( {{x^2} + 3x + 3} \right)\,\,\sqrt {x + 1} }}dx$ is equal to

$\frac{1}{{\sqrt 3 }}\,\,{\tan ^{ - 1}}\,\,\left( {\frac{x}{{\sqrt {3\left( {x + 1} \right)} }}} \right)$

$\frac{2}{{\sqrt 3 }}\,\,{\tan ^{ - 1}}\,\,\left( {\frac{x}{{\sqrt {x + 1} }}} \right)$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Medium

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Application +4 | -1

\int \frac{{x + 2}}{{\left( {{x^2} + 3x + 3} \right)\,\,\sqrt {x + 1} }}dx

is equal to

View answer and explanation

Correct option: B

\frac{2}{{\sqrt 3 }}\,\,{\tan ^{ - 1}}\,\,\left( {\frac{x}{{\sqrt {3\left( {x + 1} \right)} }}} \right)

Option B,

\frac{2}{{\sqrt 3 }}\,\,{\tan ^{ - 1}}\,\,\left( {\frac{x}{{\sqrt {3\,\,\left( {x + 1} \right)} }}} \right)

, is correct because it is the result of the given integral after applying a suitable substitution and trigonometric simplification.