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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

Add a constant to the solution

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

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