∫x/[(1-x)^3]dx Let 1-x = U =∫1/u^(3)dx =1/3∫1/u^(3)du^3 =1/3In|u^3|+c =1/3In|(1-x)^3|+c But it seems wrong,

The first step is wrong, and the next step is even more wrong. The molecule changes from X to 1-u instead of 1
So int (x / (1-x) ^ 3) DX
= int ((1-u) / u^3) d(-u)
= int (u-1) / u^3 du
= int 1/u^2 du - 1/u^3 du
= -1/u + 1/2u^2 + C.

∫1/(1+x^2)^3/2 dx

Let x = tany, DX = sec & sup2; YDY
(1+x²)^(3/2)=(1+tan²y)^(3/2)=(sec²y)^(3/2)=sec³y
The original formula = ∫ sec & sup2; Y / sec & sup3; y dy
=∫cosy dy
=siny+C
=x/√(1+x²)+C

For the integral of x ^ 3 / (x + 3) DX, thank you
Let's find another one, the integral of X + 1 / (x ^ 2-2x + 5)

x³/(x+3)
=(x³+3x²-3x²-9x+9x+27-27)/(x+3)
=x²-3x+9-27/(x+3)
So the original formula = x & sup3 / 3-3x & sup2 / 2 + 9x-27ln|x + 3| + C

∫x/[(1-x)^3]dx Let 1-x = U =∫1/u^(3)dx =1/3∫1/u^(3)du^3 =1/3In|u^3|+c =1/3In|(1-x)^3|+c But it seems wrong,