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

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

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

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

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

Solution steps

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

Apply u-substitution

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

Take the partial fraction of

\frac{1}{u ( u - 3 ) ^{2}} : \frac{1}{9 u} - \frac{1}{9 ( u - 3 )} + \frac{1}{3 ( u - 3 ) ^{2}}

= \int \frac{1}{9 u} - \frac{1}{9 ( u - 3 )} + \frac{1}{3 ( u - 3 ) ^{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{1}{9 u} d u - \int \frac{1}{9 ( u - 3 )} d u + \int \frac{1}{3 ( u - 3 ) ^{2}} d u

\int \frac{1}{9 u} d u = \frac{1}{9} ln ∣ u ∣

\int \frac{1}{9 ( u - 3 )} d u = \frac{1}{9} ln ∣ u - 3 ∣

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

= \frac{1}{9} ln ∣ u ∣ - \frac{1}{9} ln ∣ u - 3 ∣ - \frac{1}{3 ( u - 3 )}

Substitute back

u = x + 2

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

Simplify

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

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

Add a constant to the solution

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

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