What is the integral of (-7x^2+2x-1)/((x-1)^2(3x-2)) ?

The integral of (-7x^2+2x-1)/((x-1)^2(3x-2)) is 6ln|x-1|+6/(x-1)-25/3 ln|3x-2|+C

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integral of (-7x^2+2x-1)/((x-1)^2(3x-2))

\int \frac{- 7 x ^{2} + 2 x - 1}{( x - 1 ) ^{2} ( 3 x - 2 )} d x

6 ln ∣ x - 1 ∣ + \frac{6}{x - 1} - \frac{25}{3} ln ∣ 3 x - 2 ∣ + C

Solution steps

\int \frac{- 7 x ^{2} + 2 x - 1}{( x - 1 ) ^{2} ( 3 x - 2 )} d x

Take the partial fraction of

\frac{- 7 x ^{2} + 2 x - 1}{( x - 1 ) ^{2} ( 3 x - 2 )} : \frac{6}{x - 1} - \frac{6}{( x - 1 ) ^{2}} - \frac{25}{3 x - 2}

= \int \frac{6}{x - 1} - \frac{6}{( x - 1 ) ^{2}} - \frac{25}{3 x - 2} d x

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{6}{x - 1} d x - \int \frac{6}{( x - 1 ) ^{2}} d x - \int \frac{25}{3 x - 2} d x

\int \frac{6}{x - 1} d x = 6 ln ∣ x - 1 ∣

\int \frac{6}{( x - 1 ) ^{2}} d x = - \frac{6}{x - 1}

\int \frac{25}{3 x - 2} d x = \frac{25}{3} ln ∣ 3 x - 2 ∣

= 6 ln ∣ x - 1 ∣ - (- \frac{6}{x - 1}) - \frac{25}{3} ln ∣ 3 x - 2 ∣

Simplify

= 6 ln ∣ x - 1 ∣ + \frac{6}{x - 1} - \frac{25}{3} ln ∣ 3 x - 2 ∣

Add a constant to the solution

= 6 ln ∣ x - 1 ∣ + \frac{6}{x - 1} - \frac{25}{3} ln ∣ 3 x - 2 ∣ + C

What is the integral of (-7x^2+2x-1)/((x-1)^2(3x-2)) ?
The integral of (-7x^2+2x-1)/((x-1)^2(3x-2)) is 6ln|x-1|+6/(x-1)-25/3 ln|3x-2|+C