What is the integral of 9/((x+1)(x-2)^2) ?

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

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integral of 9/((x+1)(x-2)^2)

\int \frac{9}{( x + 1 ) ( x - 2 ) ^{2}} d x

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

Solution steps

\int \frac{9}{( x + 1 ) ( x - 2 ) ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \int \frac{1}{( x + 1 ) ( x - 2 ) ^{2}} d x

Apply u-substitution

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

= 9 \cdot \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}}

= 9 \cdot \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

= 9 (\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 )}

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

Substitute back

u = x + 1

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

Simplify

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

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

Add a constant to the solution

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

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