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

$3\;{({\tan ^{ - 1}}\,\,x)^2} + \,\,c$

${({\tan ^{ - 1}}\,\,x)^4} + \,\,c$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

\int \,\,\frac{{{{({{\tan }^{ - 1}}\,\,x)}^3}}}{{1\,\, + \,\,{x^2}}}\;\,\,dx

is equal to

View answer and explanation

Correct option: B

\frac{{{{({{\tan }^{ - 1}}\,\,x)}^4}}}{4}\,\, + \,\,c

Option B (

\frac{{{{({{\tan }^{ - 1}}\\ x)}^4}}}{4}\\ + \\ c

) is correct because it is the result of integrating the given expression using u-substitution.