integrate xsecx^2tanx^2
(notice: not, xsec^2xtan^x)
No, you have to substitute back. And the 2 is in the denominator !
Because when subbing y=x², dy=2x dx ---> dx=dy/(2x)
a way to check a result : Wolfram Mathematica Online Integrator
Here is what I would do :
Substituting $\displaystyle y=x^{1/4}+x$ ? Not good. The dy will have a nasty expression !
Substituting $\displaystyle y=x^{1/4}$ ? Not bad... But it would be something weird though
Another method ? Why not ?
$\displaystyle \frac{1}{x^{1/4}+x}=\frac{1}{x^{1/4} \left(1+x^{3/4}\right)}$
Now, substitue $\displaystyle y=x^{3/4}$
$\displaystyle dy=\frac 34 \cdot x^{-1/4} ~dx \implies dx=\frac 43 \cdot \frac{dy}{x^{-1/4}}$
Going back to the integral :
$\displaystyle 2 \int \frac{1}{x^{1/4} \left(1+x^{3/4}\right)}=2 \int \frac{1}{{\color{red}x^{1/4}} \left(1+y\right)} \cdot \frac 43 \cdot \frac{1}{\color{red}x^{-1/4}} ~dy$
Do you see any simplification with the red parts ??
Edit : you could have substituted $\displaystyle y=1+x^{3/4}$, but it doesn't matter, you got it