Integrate [y^2(1+y)^2]dy (using substitution?)?

Question

i tried

U=(1+y)

du=1dy

but then what? I've got [y^2(U)^2]du

Answer 1

Take y^3 = u

du = 3 y^2 dy

Th3e integral changes to [(1/3) (1+u^(1/3))^2]du

expand the square and integrate

Answer 2

∫ y^2 (1 + y)^2 dy =

no substitution is needed here; rather, expand the integrand as:

∫ y^2 (1 + 2y + y^2) dy =

∫ (y^2 + 2y^3 + y^4) dy =

and break it up into:

∫ y^2 dy + 2 ∫ y^3 dy + ∫ y^4 dy =

[y^(2+1)]/(2+1) + 2[y^(3+1)]/(3+1) + [y^(4+1)]/(4+1) + C =

(1/3)y^3 + 2(1/4)y^4 + (1/5)y^5 + C

in conclusion:

∫ y^2 (1 + y)^2 dy = (1/3)y^3 + (1/2)y^4 + (1/5)y^5 + C

I hope it helps...

bye!