Integrate tan^-1 x/x^2 dx

I know that substitution will work for this problem.
tan^-1x=u dv=x^-2 or 1/x^2
du=x/1+x^2 dx v= -1/x

-1/x tan^-1(x) - integral(-1/x)* 1/1+x^2 dx we need to use the substitution but from hear I do not know how to solve it

∫ arctan(x) / x^2 dx: => integration by parts:
u = arctan(x) , dv = x^(-2)
du = dx/(1 + x^2) , v = -x^(-1)
∫ arctan(x) / x^2 dx:
= -x^(-1) arctan(x) - ∫ -x^(-1)[1/(1 + x^2)] dx
= -1/x arctan(x) + ∫ 1/[x(1 + x^2)] dx => use partial fractions:
1/[x(1 + x^2)] = a/x + (b + cx)/(1 + x^2)
a(1 + x^2) + (b+ cx)x = 1
a + ax^2 + bx + cx^2 = 1
(a + c)x^2 + bx + a = 1
a + c = 0 => a = -c
b = 0
a = 1
c = -1
a/x + (b + cx)/(1 + x^2) = 1/x - x/(1 + x^2)
=> -1/x arctan(x) + ∫ 1/[x(1 + x^2)] dx
= -1/x arctan(x) + ∫ [1/x - x/(1 + x^2)] dx
= -1/x arctan(x) + ∫ 1/x dx - ∫ x/(1 + x^2) dx
= -1/x arctan(x) + ln(x) - ∫ x/(1 + x^2) dx => sub. u = 1 + x^2 , du = 2x dx, dx = 1/(2x) du
= -1/x arctan(x) + ln(x) - ½ ∫ 1/u du
= -1/x arctan(x) + ln(x) - ½ ln(u) => sub.back for u = 1 + x^2
= -1/x arctan(x) + ln(x) - ½ ln(1 + x^2) + C

-(ArcTan[x]/x) + Log[x/Sqrt[1 + x^2]]