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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

\int \frac{1}{9 ( x + 1 )} d x = \frac{1}{9} ln ∣ x + 1 ∣

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

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

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

Add a constant to the solution

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

y^{^{'}} = 2 y - y^{2}

\int 2 p x d x

\frac{d}{d x} ((cos (2 x))^{x})

\int_{1}^{\infty} 2 e^{- 2 x} d x

13 x^{3}

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