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

The integral of (x+3)/((x-1)(x+2)^2) is x/(3(x+2))+4/9 ln|x-1|-4/9 ln|x+2|+1/(x+2)-1/9-2/9+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

\frac{x}{3 ( x + 2 )} + \frac{4}{9} ln ∣ x - 1 ∣ - \frac{4}{9} ln ∣ x + 2 ∣ + \frac{1}{x + 2} - \frac{1}{9} - \frac{2}{9} + 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 = \frac{x}{3 ( x + 2 )} + \frac{1}{9} ln ∣ x - 1 ∣ - \frac{1}{9} ln ∣ x + 2 ∣ - \frac{1}{9} - \frac{2}{9}

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

= \frac{x}{3 ( x + 2 )} + \frac{1}{9} ln ∣ x - 1 ∣ - \frac{1}{9} ln ∣ x + 2 ∣ - \frac{1}{9} - \frac{2}{9} + \frac{1}{3} ln ∣ x - 1 ∣ - \frac{1}{3} ln ∣ x + 2 ∣ + \frac{1}{x + 2}

Simplify

\frac{x}{3 ( x + 2 )} + \frac{1}{9} ln ∣ x - 1 ∣ - \frac{1}{9} ln ∣ x + 2 ∣ - \frac{1}{9} - \frac{2}{9} + \frac{1}{3} ln ∣ x - 1 ∣ - \frac{1}{3} ln ∣ x + 2 ∣ + \frac{1}{x + 2} : \frac{x}{3 ( x + 2 )} + \frac{4}{9} ln ∣ x - 1 ∣ - \frac{4}{9} ln ∣ x + 2 ∣ + \frac{1}{x + 2} - \frac{1}{9} - \frac{2}{9}

= \frac{x}{3 ( x + 2 )} + \frac{4}{9} ln ∣ x - 1 ∣ - \frac{4}{9} ln ∣ x + 2 ∣ + \frac{1}{x + 2} - \frac{1}{9} - \frac{2}{9}

Add a constant to the solution

= \frac{x}{3 ( x + 2 )} + \frac{4}{9} ln ∣ x - 1 ∣ - \frac{4}{9} ln ∣ x + 2 ∣ + \frac{1}{x + 2} - \frac{1}{9} - \frac{2}{9} + C

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What is the integral of (x+3)/((x-1)(x+2)^2) ?
The integral of (x+3)/((x-1)(x+2)^2) is x/(3(x+2))+4/9 ln|x-1|-4/9 ln|x+2|+1/(x+2)-1/9-2/9+C