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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Expand

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

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

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

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

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

Substitute back

u = x + 1

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

Simplify

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

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

Add a constant to the solution

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

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