Verified answer

∫ [x e^x /(x + 1)²] dx =

rewrite it as a product:

∫ (x e^x) (x + 1)^(-2) dx =

let:

x e^x = u → [(1)e^x + x e^x] dx = du → (e^x + x e^x) dx = du →

e^x (1 + x) dx = du

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

[(x + 1)^(-1)] /(-1) = v → [- 1 /(x + 1)] = v

thus, integrating by parts, you get:

∫ u dv = v u - ∫ v du →

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

(x + 1) canceling out, you get:

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

[- x e^x /(x + 1)] + e^x + c =

e^x {[- x /(x + 1)] + 1} + c =

e^x [(- x + x + 1)/(x + 1)] + c

e^x [1 /(x + 1)] + c

thus, finally:

∫ [x e^x /(x + 1)²] dx = [e^x /(x + 1)] + c

I hope it helps

This problem can be solved using the formula

∫[f(x) + f '(x)] e^x dx = f(x) e^x + c

∫ xe^x/(x+1)^2 dx

= ∫[(x+1) - 1] e^x / (x+1)^2 dx

= ∫[ 1/(x+1) - 1/(x+1)^2 ] e^x dx

If 1/(x+1) = f(x), then f '(x) = - 1/(x+1)^2

=> Integral

= e^x / (x+1) + c.

For more selected problems of integration, you may refer to my free educational website of physics and mathematics listed below which I have created as a part of my hobby.

∫ xe^x/(x+1)^2 dx = ∫ (e^x) * [ x/(x+1)^2 ] dx

= ∫ (e^x) * [ 1/(x+1) + (-1)/(x+1)^2 ] dx

We have a result in IC which is

∫ e^x [f(x) + f'(x)] dx = (e^x) * f(x)

Hence

∫ xe^x/(x+1)^2 dx = (e^x) / (x+1) + c