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

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

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

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

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

Solution steps

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

Apply u-substitution

= \int \frac{( u + 1 ) ( - 5 u - 2 )}{u ^{2} ( 3 u + 2 )} d u

Take the partial fraction of

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

= \int - \frac{2}{u} - \frac{1}{u ^{2}} + \frac{1}{3 u + 2} d u

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{2}{u} d u - \int \frac{1}{u ^{2}} d u + \int \frac{1}{3 u + 2} d u

\int \frac{2}{u} d u = 2 ln ∣ u ∣

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

\int \frac{1}{3 u + 2} d u = \frac{1}{3} ln ∣ 3 u + 2 ∣

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

Substitute back

u = x - 1

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

Simplify

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

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

Add a constant to the solution

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

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