The value of $\mathop \smallint \nolimits_0^1 \,\,\frac{{{{\tan }^{ - 1}}\,\,x}}{{1 + \,\,{x^2}}}\;\,\,dx$ is

$\frac{{{{\rm{\pi }}^2}}}{{32}}$

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

The value of

\int \nolimits_0^1 \,\,\frac{{{{\tan }^{ - 1}}\,\,x}}{{1 + \,\,{x^2}}}\;\,\,dx

View answer and explanation

Correct option: B

\frac{{{\mathrm{\pi }^2}}}{{32}}

Option B (

\frac{{{\mathrm{\pi }^2}}}{{32}}

) is correct because it is the result of a u-substitution and evaluating the resulting integral.

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