Question

How do you integrate `int xtan^-1x` by integration by parts method?

Answer 1

I got:

`(x^2)/2arctanx + arctanx/2 - x/2 + C`

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Just pick the term that is easier to integrate, and use that as your `dv`.

For integration by parts, you always start with:

`\mathbf(int udv = uv - intvdu)`

So, we can pick our `u` as `arctanx`, because who knows the antiderivative of `arctanx`? Not me. However, I do know that its derivative is `1/(1+x^2)`, so we can do that.

Then, of course, now that we've done that, `dv = xdx`.

`u = arctanx, du = 1/(1+x^2)dx`
`dv = xdx, v = x^2/2`

Therefore, just follow the formula:

`color(blue)(int xarctanxdx)`

`= (x^2)/2arctanx - 1/2intx^2/(1+x^2)dx`

Now the challenge is to integrate this one. Fortunately, it's not that bad.

`=> (x^2)/2arctanx - 1/2[int (x^2 + 1 - 1)/(x^2 + 1)dx]`

`= (x^2)/2arctanx - 1/2[int dx - int 1/(x^2 + 1)dx]`

`= (x^2)/2arctanx - 1/2[x - arctanx]`

And finishing things up:

`= color(blue)((x^2)/2arctanx + arctanx/2 - x/2 + C)`