Question

How do you integrate `int x^3 sec^2 x dx` using integration by parts?

Answer 1

The first two steps are OK. After that you need polylogarithmic functions (beyond the scope of an introductory course).

Explanation:

Let `I = int x^3 sec^2 x dx`

Let `u=x^3` and `dv = sec^2x dx`.

This makes `du = 3x^2` and `v = int sec^2x dx = tanx`

So, we have

`uv-int v du = x^3 tanx - 3 int x^2 tanx dx`

To evaluate `int x^2 tanx dx` use parts again.

Let `u = x^2` and `dv = tanx dx`.

So that `du = 2x dx` and `v = int tanx dx = -ln abs cosx`

So we have

`I = x^3 tanx -3[-x^2 ln abs cosx+2 int x ln abs cosx dx]`

The last integral involves the polylogarithmic function.

I don't know enough about that to explain it. (Good luck.)