Sir please answer and specify the reason.

∫ nx^n-1/(1+x). dx = (∫ n.x^n-1.dx)/(1+x) |₀¹ – ∫ (∫n.x^n-1.dx).d_/dx(1_/(1+x)), using recerse chain rule for integration.
∫ n.x^n-1.dx = xⁿ and d_/dx(1_/(1+x)) = -1_/(1+x)², Put this above and take limits.
Lt ∫ nx^n-1/(1+x). dx = Lt (1ⁿ/(1+1) – 0/(0+1))   +    Lt ∫ xⁿ/(1+x)² .dx,      -(1) 
Now Lt xⁿ/(1+x)² =0,
(1)- Lt ∫ nx^n-1/(1+x). dx = ½