`int x\ sec^2 x\ dx`

We choose `u=x` (since it will give us a simpler `du`) and this gives us `du=dx`.

Then `dv` will be `dv=sec^2x\ dx` and integrating this gives `v=tan\ x`.

Substituting these into the Integration by Parts formula gives:

`int x\ sec^2 x\ dx =intu\ dv`

`=uv-intv\ du`

`=(x)(tan\ x)-int(tan\ x)dx`

`=x\ tan\ x-(-ln\ |cos\ x|)+K`

`=x\ tan\ x+ln\ |cos\ x|+K`

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