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

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

$3\,\,{\tan ^{ - 1}}\,\,\frac{{\sqrt {{x^2} + \,\,x\,\, + 1} }}{x}\,\, + \,\,C$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

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Analyzing +4 | -1

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

is equal to

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because the integral simplifies to this expression after substitution and integration.