The value of $\smallint \,\,{e^{{{\tan }^{ - 1}}x}}\,\,.\,\,\frac{{\left( {1 + x + {x^2}} \right)}}{{1 + {x^2}}}\,\,dx$ is

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

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

The value of

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

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because the integral can be solved by recognizing the derivative pattern within the integrand.