Question

What's the integral of `int arctan(x) dx`?

Answer 1

`intarctan(x)dx = xarctan(x) - 1/2ln(1+x^2) + C`

Explanation:

To solve `intarctan(x)dx` we will use integration by parts, that is, for functions `u` and `v`,
`intudv = uv - intvdu`.

We also need the derivative `d/dx arctan(x) = 1/(1+x^2)`
and the integral `int 1/xdx = ln|x| + C`

For our integration by parts, we will take
`u = arctan(x)` and `v = x` so
`du = 1/(1+x^2)dx` and `dv = dx`

Then, by the formula above,

`intarctan(x)dx = xarctan(x) - intx*1/(1+x^2)dx`

To solve the remaining integral, we will use `u`-substitution, with `u = 1+x^2` and `du = 2xdx`

`intx/(1+x^2)dx = 1/2int1/(1+x^2)*2xdx = 1/2int1/udu = 1/2ln|u|+C`

Substituting back gives us

`intx/(1+x^2)dx = 1/2ln|1+x^2| + C`

Finally, we plug this into the original equation to obtain

`intarctan(x)dx = xarctan(x) - 1/2ln(1+x^2) + C`

(note we do not need to use the absolute value for the natural log here as `1 + x^2 > 0" " AAx in RR`)