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

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

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

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

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Application +4 | -1

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

is equal to

View answer and explanation

Correct option: A

x\,\;{e^{{{\tan }^{ - 1}}x}}\,\, + \,\,C

Option A (

x\;{e^{{{\tan }^{ - 1}}x}}\, + \,C

) is correct because it results from a u-substitution and simplification of the integral.