Question

Evaluate the integral of arctan(1/x)

Answer 1

We will use substitution for the partial integration:
`u=arc tan ( (1)/(x)), dv = dx \\ du= (-dx)/((1+ x^(2)) x^(2) ) , v=x`
The integral becomes:=
`x arc tan x- \int {x (-dx)/((1+ x^(2)) x^(2) ) } \, dx = \\ =x arc tan x+ \int { (dx)/((1+ x^(2) )x) } \, dx`
2nd integration:
we will add and subtract x^2 to the numerator:
`\int { (1+ x^(2) - x^(2) )/((1+ x^(2) )x) } \, dx= \\ \int { (1)/(x) } \, dx- \int { (x)/(1+ x^(2) ) } \, dx = \\ ln(x) - \int { (x)/(1+ x^(2) ) } \, dx`
u-substitution:
`u=1+ x^(2) , du=2xdx, xdx= (du)/(2)`
...=
`ln(x)- (1)/(2) \int { (1)/(u) } \, du=ln(x)- (1)/(2) ln(u)= \\ ln(x)-ln( \sqrt{1+ x^(2) )} =ln \frac{x}{ \sqrt{1+ x^(2) } }`
Finally:...=
`xarctan( (1)/(x))+ln( \frac{x}{ \sqrt{1+ x^(2) } })+C`