Indefinite integral wrt x of
$$x^{2}\sqrt{0.5x+3} $$?

Question

Indefinite integral wrt x of 
$$x^{2}\sqrt{0.5x+3} $$?

anonymous · Answer

what is wrt?

anonymous · Answer

with respect to

anonymous · Answer

i think you should take couple integration by parts:
1) u=x^2  du=2x
dv=(.5x+3)^1/2; v=4/3  (2/3 x +3)^3/2

i probably one more time to reduce the power of x^2

anonymous · Answer

first integration by parts would give:
$$4/3 x^2 (1/2 x +3)^{3/2} -8/3 \int\limits_{ }^{}x*(1/2 x+3)^{3/2}dx=$$

I think one more time would do it... :)

anonymous · Answer

did you finished it ?
need help?

anonymous · Answer

Put u = 0.5x+3 (x = 2u-6) , du = 0.5 ->

Int 2(2u-6)^2 root u du

Int(8u^(5/2) -48u^(3/2)+72u^(1/2) du and sub back for u

anonymous · Answer

I got final result as (but I would check coefficients... was rushing...):
$$=4/3 *x ^{2}(1/2 x +3)^{3/2}-32/15  *x(1/2 x+3)^{5/2} +$$
$$+128/105 (1/2 x +3)^{7/2} + const$$

anonymous · Answer

Basically right, as you say coefficients are off a bit.