# U substitution question... please help !?

integral of x(x − 3)^ (5/2) dx

by making the substitution u = x − 3

im not sure on how to do this. any help would be appreciated!!

### 3 Answers

- 7 years agoBest Answer
u = x - 3

Derive implicitly

du = dx - 0

du = dx

So, x = u + 3

u = x - 3

and

du = dx

x * (x - 3)^(5/2) * dx =>

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

u^(7/2) * du + 3 * u^(5/2) * du

Integrate by using the power rule

(1/(1 + 7/2)) * u^(7/2 + 1) + 3 * (1/(1 + 5/2)) * u^(5/2 + 1) + C =>

(1/(9/2)) * u^(9/2) + 3 * (1/(7/2)) * u^(7/2) + C =>

(2/9) * u^(9/2) + (6/7) * u^(7/2) + C =>

(2/9) * (1/7) * u^(7/2) * (7 * u^(2/2) + 3 * 9 * 1) + C =>

(2/63) * (7u + 27) * u^(7/2) + C =>

(2/63) * (7 * (x - 3) + 27) * (x - 3)^(7/2) + C =>

(2/63) * (7x - 21 + 27) * (x - 3)^(7/2) + C =>

(2/63) * (7x + 6) * (x - 3)^(7/2) + C

- RaffaeleLv 77 years ago
u = x − 3

x = u + 3

dx = du

∫x(x − 3)^(5/2) dx = ∫(u + 3)u^(5/2) du =

= ∫u⋅u^(5/2) du + 3∫u^(5/2) du =

= ∫u^(7/2) du + 3∫u^(5/2) du =............... ∫uⁿ du = (1/(n + 1)) uⁿ⁺¹ + C

= (1/(1 + 7/2)) u^(1 + 7/2) + (3/(1 + 5/2)) u^(1 + 5/2) + C

= (2/9)u^(9/2) + (6/7) u^(7/2) + C =

= (2/63)(7u + 27) u^(7/2) + C =

= (2/63)(7(x - 3) + 27) (x - 3)^(7/2) + C =

= (2/63)(7x + 6) (x - 3)^(7/2) + C =

- grunfeldLv 77 years ago
int [ x ( x - 3 )^(5 / 2 ) dx ]

Using the tabular method I get :

= (2x / 7 )(x - 3 )^(7 / 2 ) - (4 / 63)(x - 3)^(9 / 2 ) + C

Source(s): my brain