how to integrate this?

Question

∫x(1-x)^3 dx using integration by parts?

Answer 1

∫x(1-x)^3 dx
= x∫(1-x)^3 dx - ∫[∫(1-x)^3 dx] (dx/dx) dx
= -x[(1-x)^4]/4 + ∫[(1-x)^4]/4 dx
= 1/4{ -x[(1-x)^4] - [(1-x)^5]/5 }
= (-1/4) (1-x)^4 [x+(1-x)/5]
= (-1/20) (1-x)^4 (1+4x)

Answer 2

let u = x^2 dv = (1-x)^3 dx
du = 2x dx v =-1/4*(1-x)^4

Integral = -1/4 x^2 (1-x)^4 +1/2integral(x (1-x)^4 dx)
now let w = x dz = (1-x)^4 dx
dw = dx z = -1/5 (1-x)^5

Integral = -1/4 x^2 (1-x)^4 -1/10 x (1-x)^5 +Integral (1/10 (1-x)^5dx)
= -1/4 x^2 (1-x)^4 -1/10 x (1-x)^5 - 1/60 (1-x)^6

Answer 3

I = ∫ x (1 - x)³ dx
I = ∫ (u )(dv/dx) dx = u v - ∫ v (du/dx) dx
where u = x and dv/dx = (1 - x) ³
du / dx = 1 and v = - (1 - x)^4 / 4
I = - [ x (1 - x)^4 ] / 4 + (1/4) ∫ (1 - x)^4 dx
I = - (1/4) [ x (1 - x)^4 ] - (1/20)(1 - x)^5 + C
I = (- 1/20)(1 - x)^4[ 5x + (1 - x)] + C
I = (- 1/20) (1 - x)^4 (4x + 1) + C

Answer 4

u = x
dv = (1-x)^3 dx

du = dx
v= -1/4(1-x)^4

integral(u dv) = uv - integral(v du)
by now, it is not difficult to finish this. ©