Question

How do you evaluate the integral `int arctansqrtx`?

Answer 1

`intarctan(sqrtx)dx=(x+1)arctan(sqrtx)-sqrtx+C`

Explanation:

`I=intarctan(sqrtx)dx`

Let `t=sqrtx`. Finding `dx` in a usable form is simpler if we first write that `t^2=x`, which then implies that `2tdt=dx`. This way we can plug in `dx` into the integral straight away. These substitutions yield:

`I=intarctan(t)(2tdt)=int2tarctan(t)dt`

Now use integration by parts. Let:

`{(u=arctan(t)" "=>" "du=1/(t^2+1)dt),(dv=2tdt" "=>" "v=t^2):}`

Then:

`I=t^2arctan(t)-intt^2/(t^2+1)dt`

Rewriting the integrand as `(t^2+1-1)/(t^2+1)=(t^2+1)/(t^2+1)-1/(t^2+1)=1-1/(t^2+1)` we see that

`I=t^2arctan(t)-(intdt-int1/(t^2+1)dt)`

Both of which are common integrals:

`I=t^2arctan(t)-t+arctan(t)+C`

Returning to `x` from `t=sqrtx`:

`I=xarctan(sqrtx)-sqrtx+arctan(sqrtx)+C`

`I=(x+1)arctan(sqrtx)-sqrtx+C`