# 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

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 .

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