There's a third way, but you'll really hate it: trig substitution

x * (1 - x)^(1/2) * dx

x = sin(t)^2

dx = 2 * sin(t) * cos(t) * dt

sin(t)^2 * (1 - sin(t)^2)^(1/2) * 2 * sin(t) * cos(t) * dt =>

2 * sin(t)^3 * cos(t) * (cos(t)^2)^(1/2) * dt =>

2 * sin(t)^3 * cos(t) * cos(t) * dt =>

2 * sin(t)^2 * cos(t)^2 * sin(t) * dt =>

2 * (1 - cos(t)^2) * cos(t)^2 * sin(t) * dt =>

2 * cos(t)^2 * sin(t) * dt - 2 * cos(t)^4 * sin(t) * dt

u = cos(t)

du = -sin(t) * dt

2 * u^2 * (-du) - 2 * u^4 * (-du) =>

2 * u^4 * du - 2 * u^2 * du

Integrate

(2/5) * u^5 - (2/3) * u^3 + C =>

(2/5) * cos(t)^5 - (2/3) * cos(t)^3 + C =>

2 * (1/5) * (1/3) * cos(t)^3 * (3 * cos(t)^2 - 5) + C =>

(2/15) * cos(t)^2 * (3 * cos(t)^2 - 5) * cos(t) + C =>

(2/15) * (1 - sin(t)^2) * (3 * (1 - sin(t)^2) - 5) * sqrt(1 - sin(t)^2) + C =>

(2/15) * (1 - x) * (3 * (1 - x) - 5) * sqrt(1 - x) + C =>

(2/15) * (1 - x) * (3 - 3x - 5) * sqrt(1 - x) + C =>

(2/15) * (1 - x) * (-2 - 3x) * sqrt(1 - x) + C =>

(-2/15) * (1 - x) * (2 + 3x) * sqrt(1 - x) + C =>

(-2/15) * (3x + 2) * (1 - x)^(3/2) + C

I = ∫[x√(1 - x)]dx, Let x = 1 - u, so that dx = -du

I = ∫[(1 – u)*(u)^(1/2)]*-du = ∫[u^(3/2) – u^(1/2)]du

I = (2/5)u^(5/2) – (2/3)u^(3/2) =

I = (2/15)(3u – 5)u^(3/2) + C

I = (2/15)(3u – 5)u^(3/2) + C

I = -(2/15)(3x + 2)(1 – x)^(3/2) + C

Easier than integrating by parts. That starts with

I = x*∫√(1 - x)dx - ∫[(-2/3)*(1 - x)^(3/2) dx

The second term becomes –(4/15)(1 – x)^(3/2)

and the algebra leads to same result

I = -(2/15)(3x + 2)(1 – x)^(3/2) + C

The answers will not differ except in appearance.

(1) u = 1-x, x = 1 - u, the integral of [x(1-x)^(1/2) dx] would be

the integral of -(1 - u)*[u^(1/2)] du

= -(2/3)u^(3/2) + (2/5)u^(5/2) + C

= -(2/3)(1-x)^(3/2) + (2/5)(1-x)^(5/2) + C. Call this f(x).

(2) Now examine Iggy Rocko's result from integrating by parts. He says

-(2/3)x(1-x)^(3/2) - (4/15)(1-x)^(5/2) + C. Call this g(x).

Note that f(0)= -2/3 + 2/5 = -4/15 = g(0), and

f(-1) = -(2/3)2^(3/2) + (2/5)2^(5/2) = -(2/3-4/5)2^(3/2) = (2/15)*2^(3/2);

g(-1) = (2/3)2^(3/2) - (8/15)2^(3/2) = (2/15)*2^(3/2) = f(-1).

And similarly, for any value you care to plug in, you will find f(x) = g(x).

17$ microwave 17 phone book %of 76zoo animals. What is with this face maths crap it's makes me so angry.

integration by parts:

Let u = x and dv = (1 - x)^(1/2) dx

∫x(1 - x)^(1/2) dx =

x * -2(1 - x)^(3/2)/3 - ∫(-2(1 - x)^(3/2)/3 * 1) dx =

-2x(1 - x)^(3/2)/3 - 4(1 - x)^(5/2)/15 + c