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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Expand

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

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

\int \frac{3}{u ^{2} ( u - 3 ) ^{2}} d u = 3 (\frac{2}{27} ln ∣ u ∣ - \frac{1}{9 u} - \frac{2}{27} ln ∣ u - 3 ∣ - \frac{1}{9 ( u - 3 )})

= 2 (\frac{1}{9} ln ∣ u ∣ - \frac{1}{9} ln ∣ u - 3 ∣ - \frac{1}{3 ( u - 3 )}) - 3 (\frac{2}{27} ln ∣ u ∣ - \frac{1}{9 u} - \frac{2}{27} ln ∣ u - 3 ∣ - \frac{1}{9 ( u - 3 )})

Substitute back

u = x + 1

= 2 (\frac{1}{9} ln ∣ x + 1 ∣ - \frac{1}{9} ln ∣ x + 1 - 3 ∣ - \frac{1}{3 ( x + 1 - 3 )}) - 3 (\frac{2}{27} ln ∣ x + 1 ∣ - \frac{1}{9 ( x + 1 )} - \frac{2}{27} ln ∣ x + 1 - 3 ∣ - \frac{1}{9 ( x + 1 - 3 )})

Simplify

2 (\frac{1}{9} ln ∣ x + 1 ∣ - \frac{1}{9} ln ∣ x + 1 - 3 ∣ - \frac{1}{3 ( x + 1 - 3 )}) - 3 (\frac{2}{27} ln ∣ x + 1 ∣ - \frac{1}{9 ( x + 1 )} - \frac{2}{27} ln ∣ x + 1 - 3 ∣ - \frac{1}{9 ( x + 1 - 3 )}) : - \frac{1}{3 ( x - 2 )} + \frac{1}{3 ( x + 1 )}

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

Add a constant to the solution

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

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