# Evaluate the definite integral. `int_0^3 x^2sqrt(x+1)dx` Not to sure how to go about this question, i wanted to use u subsitution where i would put u=x+1 then du=dx, the only trouble is that...

Evaluate the definite integral.

`int_0^3 x^2sqrt(x+1)dx`

Not to sure how to go about this question, i wanted to use u subsitution where i would put u=x+1 then du=dx, the only trouble is that x^2 is still left in the equation. So then i was wondering if it was possible to expand it before evaluating the integral? Help is apprecaited! Thank you in advance!

### 1 Answer | Add Yours

`int_0^3 x^2sqrt(x+1)dx=int_0^3 x^2(x+1)^(1/2)dx`

Apply integration by parts. The formula is `int udv=uv-int vdu` .

`u=x^2` and `dv=int(x+1)^(1/2)dx`

`du=2xdx` ` v=2/3(x+1)^(3/2)`

Substitute u,v,and du to the formula.

`int x^2(x+1)^(1/2)dx=2/3x^2(x+1)^(3/2)-4/3 int x(x+1)^(3/2)dx`

To evaluate the integral part, use integration by parts again. Let,

`u=x ` and `dv=int(x+1)^(3/2)dx`

`du=dx` `v=2/5(x+1)^(5/2)`

So,

`int x^2(x+1)^(1/2)dx=2/3x^2(x+1)^(3/2)-4/3(2/5x(x+1)^(5/2)-2/5int (x+1)^(5/2)dx)`

`=2/3x^2(x+1)^(3/2)-4/3(2/5x(x+1)^(5/2)-2/5*2/7(x+1)^(7/2))`

`=2/3x^2(x+1)^(3/2)-8/15x(x+1)^(5/2)+16/105(x+1)^(7/2)`

Then, evaluate the limits of the integral.

`int_0^3 x^2(x+1)^(1/2)dx =2/3x^2(x+1)^(3/2)-8/15x(x+1)^(5/2)+16/105(x+1)^(7/2)` `|_0^3`

`=[2/3*3^2*4^(3/2)-8/15*3*4^(5/2)+16/105*4^(7/2)]-[0-0+16/105]`

`=[6*2^3-24/5*2^5+16/105*2^7]-16/105`

`= 48-768/15+2048/105-16/105`

`=48-768/15+2032/105`

`=1696/105`

**Hence, `int_0^3 x^2sqrt(x+1)dx=1696/105` .**