Question

How do you integrate `int x/sqrt(2+3x)` by parts?

Answer 1

`int x/sqrt(2+3x)dx = 2/27sqrt(2+3x)(3x - 4) + C`

Explanation:

You would not integrate this integral By Parts, instead you would use a substitution ;

Let `u = 2+3x => x=1/3(u-2)`
And, `(du)/dx = 3 => 1/3int ...du = int ... dx`

Applying this substitution to the problem gives us:

`int x/sqrt(2+3x)dx = int (1/3(u-2))/sqrt(u)(1/3)du`
`:. int x/sqrt(2+3x)dx = 1/9 int (u-2)u^(-1/2)du`
`:. int x/sqrt(2+3x)dx = 1/9 int u^(1/2)-2u^(-1/2)du`
`:. int x/sqrt(2+3x)dx = 1/9 {u^(3/2)/(3/2) - (2u^(1/2))/(1/2)} + C`
`:. int x/sqrt(2+3x)dx = 1/9 u^(1/2){2/3u - 4} + C`
`:. int x/sqrt(2+3x)dx = 1/9 u^(1/2){2/3(u - 6)} + C`
`:. int x/sqrt(2+3x)dx = 2/27 u^(1/2)(u - 6) + C`

Substituting for `u` gives is;

`int x/sqrt(2+3x)dx = 2/27sqrt(2+3x)(2+3x - 6) + C`
`:. int x/sqrt(2+3x)dx = 2/27sqrt(2+3x)(3x - 4) + C`

Answer 2

Please see below.

Explanation:

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

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

This make `du = dx` and

`v = int (2+3x)^(-1/2) dx = 2/3 (2+3x)^(1/2)` (by substitution)

`I = uv-int v du`

`= 2/9x(2+3x)^(1/2) - 2/3 int (2+3x)^(1/2) dx`

`= 2/3x(2+3x)^(1/2) - 2/3 [ 2/9(2+3x)^(3/2) ] +C` (by substitution)

Clean up the algebra as desired.