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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u + 1}{u ^{2} ( u - 1 )} d u

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

Take the partial fraction of

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

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

\int \frac{2}{u} d u = 2 ln ∣ u ∣

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

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

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

Substitute back

u = x - 1

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

Simplify

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

Add a constant to the solution

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

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