Question

How do you find the antiderivative of `int (arctan(sqrtx)) dx`?

Answer 1

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

Explanation:

`I=intarctan(sqrtx)dx`

Integration by parts takes the form `intudv=uv-intvdu`. So, for `intarctan(sqrtx)dx`, we should let `u=arctan(sqrtx)` and `dv=dx`.

Differentiating `u` gives `(du)/dx=1/(1+(sqrtx)^2)*d/dxsqrtx` or `du=1/(2sqrtx(1+x))dx`. From `dv=dx` we integrate to show that `v=x`. Thus:

`I=uv-intvdu`

`color(white)(I)=xarctan(sqrtx)-intx/(2sqrtx(1+x))dx`

`color(white)(I)=xarctan(sqrtx)-1/2intsqrtx/(1+x)dx`

Letting `t=sqrtx` implies that `t^2=x` and furthermore, that `2tdt=dx` through differentiating. Thus:

`I=xarctan(sqrtx)-1/2intt/(1+t^2)(2tdt)`

`color(white)I=xarctan(sqrtx)-intt^2/(1+t^2)dt`

`color(white)I=xarctan(sqrtx)-int(1+t^2-1)/(1+t^2)dt`

`color(white)I=xarctan(sqrtx)-int(1+t^2)/(1+t^2)dt+int1/(1+t^2)dt`

`color(white)I=xarctan(sqrtx)-intdt+arctan(t)`

`color(white)I=xarctan(sqrtx)-t+arctan(t)`

Since `t=sqrtx`:

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

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