How do you integrate $$\int x^3\operatorname{sech}^2(x^2)\,\mathrm{d}x\:?$$
I've tried integration by parts, but then I find myself left with having to integrate $\operatorname{sech}(x^2)^2$ if i let it equal to $dv/dx$ (and I don't know how to work it out if i let $dv = x^3$ instead). And I don't think you can use derivative substitution considering you can't write $x^3$ in terms of the derivative of $x^2$. So how would you solve it?
If you do $x^2=u$ and $2x\,\mathrm dx=\mathrm dy$, then your integral becomes$$\frac12\int y\operatorname{sech}^2(y)\,\mathrm dy.$$Now, use integration by parts:\begin{align}\int y\operatorname{sech}^2(y)\,\mathrm dy&=y\tanh(y)-\int\tanh(y)\,\mathrm dy\\&=y\tanh(y)-\log\bigl(\cosh(y)\bigr)\\&=x^2\tanh\left(x^2\right)-\log\left(\cosh\left(x^2\right)\right).\end{align}