JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

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Application +4 | -1

The value of

\int {\frac{{1 + x + {x^2}}}{{1 + {x^2}}}{e^{{{\tan }^{ - 1}}x}}dx}

is equal to

View answer and explanation

Correct option: C

x{e^{{{\tan }^{ - 1}}x}} + c

Option C (

x{e^{{{\tan }^{ - 1}}x}} + c

) is correct because after substituting

t = \tan^{-1}x

and integrating by parts, the integral evaluates to this expression.

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