Question

How do you use substitution to integrate x/sqrt(3x-2), help ?)

Answer 1

Substitute `3x-2=u^2`, to get

`int x/sqrt{3x-2} dx=2/27(3x-2)^{3/2} +4/9sqrt{3x-2}+C`

Explanation:

Try `3x-2=u^2`, so that `3dx = 2udu`. Then

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

Answer 2

Given:

`int x/sqrt(3x-2) dx`

Let `u = 3x-2` then `du = 3dx` which can be written as `dx = 1/3du`. Substituting, the integrand becomes:

`int x/sqrt(3x-2) dx = 1/3 int x/sqrtu du`

We need an equation for x in terms of u:

`x = (u+2)/3`

`int x/sqrt(3x-2) dx = 1/9 int (u+2)/sqrtu du`

Perform the division:

`int x/sqrt(3x-2) dx = 1/9 int sqrtu+2/sqrtu du`

Separate into two integrals:

`int x/sqrt(3x-2) dx = 1/9 int sqrtu du+ 2/9 int 1/sqrtu du`

Write the square roots as a power:

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

Integrate using the power rule `int u^n du = 1/(n+1)u^(n+1)`

`int x/sqrt(3x-2) dx = 2/27 u^(3/2)+ 4/9 u^(1/2)+ C`

Reverse the substitution for u:

`int x/sqrt(3x-2) dx = 2/27 (3x-2)^(3/2)+ 4/9 (3x-2)^(1/2)+ C`