What is the integral of (x-3)/((3x-2)^2) ?

The integral of (x-3)/((3x-2)^2) is -x/(3(3x-2))+1/9 ln|3x-2|+1/(3x-2)+C

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integral of (x-3)/((3x-2)^2)

\int \frac{x - 3}{( 3 x - 2 ) ^{2}} d x

- \frac{x}{3 ( 3 x - 2 )} + \frac{1}{9} ln ∣ 3 x - 2 ∣ + \frac{1}{3 x - 2} + C

Solution steps

\int \frac{x - 3}{( 3 x - 2 ) ^{2}} d x

Expand

\frac{x - 3}{( 3 x - 2 ) ^{2}} : \frac{x}{( 3 x - 2 ) ^{2}} - \frac{3}{( 3 x - 2 ) ^{2}}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{x}{( 3 x - 2 ) ^{2}} d x - \int \frac{3}{( 3 x - 2 ) ^{2}} d x

\int \frac{x}{( 3 x - 2 ) ^{2}} d x = - \frac{x}{3 ( 3 x - 2 )} + \frac{1}{9} ln ∣ 3 x - 2 ∣

\int \frac{3}{( 3 x - 2 ) ^{2}} d x = - \frac{1}{3 x - 2}

= - \frac{x}{3 ( 3 x - 2 )} + \frac{1}{9} ln ∣ 3 x - 2 ∣ - (- \frac{1}{3 x - 2})

Simplify

= - \frac{x}{3 ( 3 x - 2 )} + \frac{1}{9} ln ∣ 3 x - 2 ∣ + \frac{1}{3 x - 2}

Add a constant to the solution

= - \frac{x}{3 ( 3 x - 2 )} + \frac{1}{9} ln ∣ 3 x - 2 ∣ + \frac{1}{3 x - 2} + C

What is the integral of (x-3)/((3x-2)^2) ?
The integral of (x-3)/((3x-2)^2) is -x/(3(3x-2))+1/9 ln|3x-2|+1/(3x-2)+C