Question

How do you find the integral of `xarctanxdx`?

Answer 1

You can see that there is one component that you can easily integrate, and one component that you could feasibly differentiate.

Although you might already know `int arctanxdx`, I will assume that you don't. Instead, I will assume that you know `d/(dx)[arctanx] = 1/(1+x^2)`.

When I see this, I see the following:

`int xarctanxdx = int udv`

which suggests Integration by Parts.

`int udv = uv - int vdu`

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

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

(With the integral here, one might be tempted to use Partial Fraction Decomposition, but there is an easier way.)

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

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

`= 1/2(x - arctanx)`

Thus, the overall integral gives:

`int xarctanxdx = x^2/2 arctanx - 1/2(x - arctanx)`

`= color(blue)(1/2[(x^2 + 1)arctanx - x] + C)`