`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:

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`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` .**