What is the integral of (3x-5x^2)/((3x-1)(x-1)^2) ?

The integral of (3x-5x^2)/((3x-1)(x-1)^2) is 1/3 ln|3x-1|-2ln|x-1|+1/(x-1)+C

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integral of (3x-5x^2)/((3x-1)(x-1)^2)

\int \frac{3 x - 5 x ^{2}}{( 3 x - 1 ) ( x - 1 ) ^{2}} d x

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

Solution steps

\int \frac{3 x - 5 x ^{2}}{( 3 x - 1 ) ( x - 1 ) ^{2}} d x

Take the partial fraction of

\frac{3 x - 5 x ^{2}}{( 3 x - 1 ) ( x - 1 ) ^{2}} : \frac{1}{3 x - 1} - \frac{2}{x - 1} - \frac{1}{( x - 1 ) ^{2}}

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

\int \frac{1}{3 x - 1} d x = \frac{1}{3} ln ∣ 3 x - 1 ∣

\int \frac{2}{x - 1} d x = 2 ln ∣ x - 1 ∣

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

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

Simplify

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

Add a constant to the solution

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

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What is the integral of (3x-5x^2)/((3x-1)(x-1)^2) ?
The integral of (3x-5x^2)/((3x-1)(x-1)^2) is 1/3 ln|3x-1|-2ln|x-1|+1/(x-1)+C