x = tan(u) dx = sec^2(u) du 

∫ (x * arctan(x) ) / ( (1+x^2)^2 ) dx 

= ∫ (tan(u) * u * sec^2(u) du) / ( (1 + tan^2(u))^2 ) 

= ∫ (tan(u) * u * sec^2(u) du) / ( sec^4(u) ) 

= ∫ (tan(u) * u du) / ( sec^2(u) ) 

= ∫ sin(u) cos(u) u du 

use sin(2a) = 2sin(a)cos(a) identity 

= (1/2)∫u sin(2u) u du 

integration by parts 

= -(1/4)u cos(2u) + (1/4)∫cos(2u) du 

= -(1/4)u cos(2u) + (1/8)sin(2u) 

= -(1/4)u (2cos^2(u) - 1) + (1/8)(2sin(u) cos(u)) 

rewrite sin and cos in terms of tan 

cos^2 = 1 / (1 + tan^2) - is readily derived 

and sin cos = tan cos^2 = tan / (1 + tan^2) 

so continuing 

integral = -(1/4)u (2 / (1 + tan^2(u)) - 1) + (1/8)(2tan(u) / (1 + tan^2(u))) 

back substitute 

= (-1/4)arctan(x)(2 / (1 + x^2) - 1) + (1/8)(2 x / (1 + x^2)) 

= x /(4 (1 + x^2)) -arctan(x) (1-x^2) /(4 (1 + x^2)) 

= (1/4)(x - (1-x^2) arctan(x)) / (1 + x^2) + c

∫ arctanx [x /(1 + x²)²] dx = 

let: 

arctanx = u → [1/(1 + x²)] dx = du 

[x /(1 + x²)²] dx = dv (divide and multiply by 2) → (1/2)[2x dx/(1 + x²)²] = dv → (1/2)[d(1 + x²)/(1 + x²)²] = dv → (1/2)(1 + x²)^(- 2) d(1 + x²) = dv → (1/2)[1/(- 2+1)](1 + x²)^(- 2+1) = v → (1/2)[1/(- 1)](1 + x²)^(- 1) = v → (-1/2)[1/(1 + x²)] = v 

integrating by parts (∫ u dv = v u - ∫ v du), you get: 

∫ arctanx [x /(1 + x²)²] dx = (-1/2)[1/(1 + x²)] arctanx - ∫ (-1/2)[1/(1 + x²)] [1/(1 + x²)] dx = 

(-1/2)[arctanx/(1 + x²)] + (1/2) ∫ [1/(1 + x²)²] dx (#) 

let us evaluate ∫ [1/(1 + x²)²] dx, adding and subtracting x² on the top: 

∫ [1/(1 + x²)²] dx = ∫ {[(1 + x²) - x²]/(1 + x²)²} dx 

break it up into: 

∫ [(1 + x²)/(1 + x²)²] dx - ∫ [x²/(1 + x²)²] dx = 

simplifying into: 

∫ [1 /(1 + x²)] dx - ∫ [x²/(1 + x²)²] dx = 

rewrite the second integrand as: 

∫ [1 /(1 + x²)] dx - ∫ x [x/(1 + x²)²] dx = 

let: 

x = u → dx = du 

[x/(1 + x²)²] dx = dv → (1/2)[2x dx/(1 + x²)²] = dv → (1/2)(1 + x²)^(- 2) d(1 + x²) = dv → (1/2)[1/(- 2+1)](1 + x²)^(- 2+1) = v → (1/2)[1/(- 1)](1 + x²)^(- 1) = v → (-1/2)[1/(1 + x²)] = v 

integrating the second integral by parts (∫ u dv = v u - ∫ v du), you get: 

∫ [1 /(1 + x²)] dx - { ∫ x [x/(1 + x²)²] dx} = 

∫ [1 /(1 + x²)] dx - {(-1/2)[1/(1 + x²)] x - ∫ (-1/2)[1/(1 + x²)] dx} = 

∫ [1 /(1 + x²)] dx - {(-1/2)[x/(1 + x²)] + (1/2) ∫ [1/(1 + x²)] dx} = 

∫ [1 /(1 + x²)] dx + (1/2)[x /(1 + x²)] - (1/2) ∫ [1/(1 + x²)] dx = 

(1/2) ∫ [1 /(1 + x²)] dx + (1/2)[x /(1 + x²)] = 

(1/2)arctanx + (1/2)[x /(1 + x²)] + C 

thus, summing up: 

∫ [1/(1 + x²)²] dx = (1/2)arctanx + (1/2)[x /(1 + x²)] + C 

plug this into the above (#) expression, yielding: 

(-1/2)[arctanx/(1 + x²)] + (1/2) ∫ [1/(1 + x²)²] dx = (-1/2)[arctanx/(1 + x²)] + (1/2) {(1/2)arctanx + (1/2)[x /(1 + x²)]} + C = 

(-1/2)[arctanx/(1 + x²)] + (1/4)arctanx + (1/4)[x /(1 + x²)] + C 

thus, in conclusion: 

∫ [(x arctanx) /(1 + x²)²] dx = (-1/2)[arctanx/(1 + x²)] + (1/4)arctanx + (1/4)[x /(1 + x²)] + C