Question

help with integrals

Answer 1

\[\int\limits_{}^{}\frac{ \tan^{-1} (x)^6 }{ 1+x^2 }\]

Answer 2

Hey there :)
Recall this derivative,\[\large\rm \frac{d}{dx}\arctan x=\frac{1}{1+x^2}\]So if you rewrite your integral like this,\[\large\rm \int\limits\frac{(\arctan x)^6}{1+x^2}dx\qquad=\qquad \int\limits (\arctan x)^6\left(\frac{1}{1+x^2}dx\right)\]You might be able to see the u-substitution.

Answer 3

Yes no maybe? :)