Answer
$ -\dfrac{1}{2} \ln |\cos (x^2+1) |+C$
Work Step by Step
Our aim is to solve the integral $ \int x \tan (x^2+1)$
Let us consider that $a =x^2+1$ and $dx=\dfrac{da}{2x}$
Now, $\int x \tan (x^2+1) = \int x \tan a \times (\dfrac{1}{2x}) (da)$
or, $= \dfrac{1}{2} \int \tan a da $
or, $=-\dfrac{1}{2} \ln |\cos a |+C$
or, $ =-\dfrac{1}{2} \ln |\cos (x^2+1) |+C$
