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

The integral of (x+3)/((x-1)(x-2)^2) is 4ln|x-1|-4ln|x-2|-5/(x-2)+C

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

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

4 ln ∣ x - 1 ∣ - 4 ln ∣ x - 2 ∣ - \frac{5}{x - 2} + C

Solution steps

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

Expand

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

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

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

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

Simplify

ln ∣ x - 1 ∣ - ln ∣ x - 2 ∣ - \frac{2}{x - 2} + 3 (ln ∣ x - 1 ∣ - ln ∣ x - 2 ∣ - \frac{1}{x - 2}) : 4 ln ∣ x - 1 ∣ - 4 ln ∣ x - 2 ∣ - \frac{5}{x - 2}

= 4 ln ∣ x - 1 ∣ - 4 ln ∣ x - 2 ∣ - \frac{5}{x - 2}

Add a constant to the solution

= 4 ln ∣ x - 1 ∣ - 4 ln ∣ x - 2 ∣ - \frac{5}{x - 2} + C

What is the integral of (x+3)/((x-1)(x-2)^2) ?
The integral of (x+3)/((x-1)(x-2)^2) is 4ln|x-1|-4ln|x-2|-5/(x-2)+C