Question

How do you integrate `int xsqrt(x+2)dx`?

Answer 1

`int xsqrt(x+2) dx = {2(x+2)^(3/2)(3x-4)}/15 + C`

Explanation:

Let `= int xsqrt(x+2) dx`

We can integrate this easily by substitution:

Let `u = x+2 => (du)/dx=1`, or `int ... du = int ... dx`

Applying the substitution we have:

`I = int (u-2)sqrtu du`
`:. I = int (u-2)u^(1/2) du`
`:. I = int u^(3/2) - 2u^(1/2) du`
`:. I = u^(5/2)/(5/2) - 2u^(3/2)/(3/2) + C`
`:. I = 2/5u^(5/2) - 4/3u^(3/2) + C`
`:. I = 2/5(x+2)^(5/2) - 4/3(x+2)^(3/2) + C`

We can put simplify by putting over a common denominator and factoring out `(x+2)^(3/2)`

`:.I = (6(x+2)^(5/2) - 20(x+2)^(3/2))/15 + C`
`:.I = (x+2)^(3/2){(6(x+2) - 20)}/15 + C`
`:.I = (x+2)^(3/2){(6x+12 - 20)}/15 + C`
`:.I = (x+2)^(3/2){(6x-8)}/15 + C`
`:.I = {2(x+2)^(3/2)(3x-4)}/15 + C`