Question

Integrate `int_1 ^sqrt3 arctan(1/x)dx`?

Answer 1

`int_(1)^(sqrt(3)) \ arctan(1/x) \ dx = (sqrt(3)pi)/6 + 1/2 ln2 - pi/4 ~~ 0.468075 ...`

Explanation:

We seek:

`I = int_(1)^(sqrt(3)) \ arctan(1/x) \ dx`

We can apply Integration By Parts

Let `{ (u,=arctan(1/x), => (du)/dx,=1/(1+1/x^2)(-1/x^2)=-1/(x^2+1)), ((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 \ (arctan(1/x))(1) \ dx = (arctan(1/x))(x) - int \ (x)(-1/(x^2+1))) \ dx`

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

And so:

`I = [xarctan(1/x) + 1/2 ln (x^2+1) ]_(1)^(sqrt(3))`

`\ \ = (sqrt(3)arctan(1/sqrt(3)) + 1/2 ln (3+1) ) - (arctan(1) + 1/2 ln (1+1) )`

`\ \ = (sqrt(3)pi/6 + 1/2 ln4 ) - (pi/4 + 1/2 ln2)`

`\ \ = sqrt(3)pi/6 + 1/2 2ln2 - pi/4 - 1/2 ln2`

`\ \ = (sqrt(3)pi)/6 + 1/2 ln2 - pi/4`

`\ \ ~~ 0.468075 ...`