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

The integral of (2x+1)/((x+1)(x-2)^2) is -1/9 ln|x+1|+1/9 ln|x-2|-5/(3(x-2))+C

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

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

- \frac{1}{9} ln ∣ x + 1 ∣ + \frac{1}{9} ln ∣ x - 2 ∣ - \frac{5}{3 ( x - 2 )} + C

Solution steps

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

Expand

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

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

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

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

Simplify

2 (- \frac{1}{9} ln ∣ x + 1 ∣ + \frac{1}{9} ln ∣ x - 2 ∣ - \frac{2}{3 ( x - 2 )}) + \frac{1}{9} ln ∣ x + 1 ∣ - \frac{1}{9} ln ∣ x - 2 ∣ - \frac{1}{3 ( x - 2 )} : - \frac{1}{9} ln ∣ x + 1 ∣ + \frac{1}{9} ln ∣ x - 2 ∣ - \frac{5}{3 ( x - 2 )}

= - \frac{1}{9} ln ∣ x + 1 ∣ + \frac{1}{9} ln ∣ x - 2 ∣ - \frac{5}{3 ( x - 2 )}

Add a constant to the solution

= - \frac{1}{9} ln ∣ x + 1 ∣ + \frac{1}{9} ln ∣ x - 2 ∣ - \frac{5}{3 ( x - 2 )} + C

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What is the integral of (2x+1)/((x+1)(x-2)^2) ?
The integral of (2x+1)/((x+1)(x-2)^2) is -1/9 ln|x+1|+1/9 ln|x-2|-5/(3(x-2))+C