# Find the Indefinite Integral using integration by substitution: `int(2x)/(3x+1) dx`this is what I have: u=3x+1 and x=(u-1)/3 then du/dx = 3 and dx= 1/3 du S2(1/3)(u-1)/u (1/3)du 2/9 S (u-1)/u...

Find the Indefinite Integral using integration by substitution: `int(2x)/(3x+1) dx`

this is what I have: u=3x+1 and x=(u-1)/3 then du/dx = 3 and dx= 1/3 du

S2(1/3)(u-1)/u (1/3)du

2/9 S (u-1)/u du

2/9 S (1-1)/u du

2/9 S u-lnu

2/9(3x+1-ln|3x+1|)+C

Thank you for your assistance

### 1 Answer | Add Yours

`int (2x)/(3x+1)dx`

Express the denominator with one term only. So, using u-substitution, let the denominator be

`u=3x+1`

Differentiate u with respect to x.

`du=3dx`

`dx=(du)/3`

Since the numerator has a variable x, from `u=3x+1` , solve for x.

`u=3x+1`

`u-1=3x`

`1/3(u-1)=x`

Replacing the x in the integrand with u yields:

`int( 2(1/3u-1))/u *(du)/3=2/9int (u-1)/u du`

Express the integrand as two fractions.

`=2/9 int (u/u-1/u) du = 2/9 int (1-1/u)du= 2/9 int du - 2/9int (du)/u`

`= 2/9u - 2/9 ln u + C`

To return to the original variable, substitute u= 3x +1.

`=2/9 (3x+1) - 2/9 ln(3x+1) + C`

`=2/3x+2/9-2/9 ln(3x+1) + C`

**Hence,` int (2x)/(3x+1) dx= 2/3x+2/9-2/9ln(3x+1)+C` .**