Question

What is `int xarctan(x^2) dx`?

Answer 1

`intxarctan(x^2)dx=1/2x^2arctan(x^2)-1/4ln(1+x^4)+C`

Explanation:

First, we can perform substitution with the `x^2` and `x`,

Let `t=x^2` so that `dt=2xdx`.

`intxarctan(x^2)=1/2int2xarctan(x^2)dx=1/2intarctan(t)dt`

Now, we will use integration by parts. Don't be fooled by the fact that there's only the arctangent function in the integrand--the `dv` value can be `1`.

As a reminder, integration by parts takes the form:

`intudv=uv-intvdu`

Here, `u=arctan(t)` and `dv=dt`. These imply that `du=1/(1+t^2)dt` and `v=t`.

Thus

`1/2intarctan(t)dt=1/2[tarctan(t)-intt/(1+t^2)dt]`

For this second integral, use substitution again.

Let `s=1+t^2` and `ds=2tdt`.

`intt/(1+t^2)dt=1/2int(2t)/(1+t^2)dt=1/2int1/sds=1/2ln(abss)+C`

Thus,

`1/2intarctan(t)dt=1/2[tarctan(t)-1/2ln(abss)]+C`

Since `s=1+t^2`,

`1/2intarctan(t)dt=1/2tarctan(t)-1/4ln(abs(1+t^2))+C`

Now since `t=x^2`:

`1/2intarctan(t)dt=1/2x^2arctan(x^2)-1/4ln(1+x^4)+C`

And `1/2intarctan(t)dt=intxarctan(x^2)dx`, so

`intxarctan(x^2)dx=1/2x^2arctan(x^2)-1/4ln(1+x^4)+C`