Question

INTEGRATION arctanx ?

Answer 1

`int \ arctan x \ dx = xarctan x - 1/2 ln(x^2+1) + C`

Explanation:

We seek:

`I = int \ arctan x \ dx`

We can apply Integration By Parts

Let `{ (u,=arctanx, => (du)/dx,=1/(1+x^2)), ((dv)/dx,=1, => v,=x ) :}`

Then plugging into the IBP formula:

`int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx`

gives us

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

`:. int \ arctan x \ dx = xarctan x - 1/2 \ int \ (2x)/(x^2+1) \ dx`
`" " = xarctan x - 1/2 ln|x^2+1| + C`

Noting that `x^2+1 gt 0 AA x in RR` we can write:

And so:

`int \ arctan x \ dx = xarctan x - 1/2 ln(x^2+1) + C`