integral of {x(x+1)^1/2}...please help me...?

Answer 1

let u = x + 1
so x = u - 1
and du=dx

∫x(x+1)^1/2 dx becomes
(u-1) * u^(1/2) du

Distribute
∫[u^(3/2) - u^1/2]

Integrate.
(2/5)u^(5/2) - (2/3)u^(3/2) + C

Put it back in terms of x using u = x+1
(2/5)(x+1)^(5/2) - (2/3)(x+1)^(3/2) + C

So,
∫x(x+1)^1/2 dx = (2/5)(x+1)^(5/2) - (2/3)(x+1)^(3/2) + C

Note: CPUcate's method is equally valid and the functions from both methods are the same, just expressed differently.

Answer 2

use integration by parts
let u = x . . . . . . . dv =(x+1) ^(1/2)
du = 1 . . . . . . . . . v = 2/3 ( x+1 )^(3/2)

integ [ x (x+1)^(1/2) ] = u v - integ v du
. . . . = 2/3 x (x+1)^(3/2) - integ 2/3 ( x+1 )^(3/2) du
. . . . = 2/3 x (x+1)^(3/2) - 4/15 ( x + 1 )^(5/2) + c

Answer 3

start with t=√(x+1), hence x=t^2-1, dx=2t*dt;

♠ thus y(t)*dt = (t^2-1)*t*2t*dt =2dt*t^4 –2dt*t^2;
Y(t)= (2/5)*t^5 –(2/3)*t^3; hence
Y(x)= (2/5) √(x+1)^5 –(2/3) √(x+1)^3 +C;

Answer 4

{1/12x^2(2x+3)}