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

The integral of (x^3+2x-1)/((x^2-x-2)^2) is 7/27 ln|x+1|+4/(9(x+1))+20/27 ln|x-2|-(11)/(9(x-2))+C

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

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

\frac{7}{27} ln ∣ x + 1 ∣ + \frac{4}{9 ( x + 1 )} + \frac{20}{27} ln ∣ x - 2 ∣ - \frac{11}{9 ( x - 2 )} + C

Solution steps

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

Take the partial fraction of

\frac{x ^{3} + 2 x - 1}{( x ^{2} - x - 2 ) ^{2}} : \frac{7}{27 ( x + 1 )} - \frac{4}{9 ( x + 1 ) ^{2}} + \frac{20}{27 ( x - 2 )} + \frac{11}{9 ( x - 2 ) ^{2}}

= \int \frac{7}{27 ( x + 1 )} - \frac{4}{9 ( x + 1 ) ^{2}} + \frac{20}{27 ( x - 2 )} + \frac{11}{9 ( 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{7}{27 ( x + 1 )} d x - \int \frac{4}{9 ( x + 1 ) ^{2}} d x + \int \frac{20}{27 ( x - 2 )} d x + \int \frac{11}{9 ( x - 2 ) ^{2}} d x

\int \frac{7}{27 ( x + 1 )} d x = \frac{7}{27} ln ∣ x + 1 ∣

\int \frac{4}{9 ( x + 1 ) ^{2}} d x = - \frac{4}{9 ( x + 1 )}

\int \frac{20}{27 ( x - 2 )} d x = \frac{20}{27} ln ∣ x - 2 ∣

\int \frac{11}{9 ( x - 2 ) ^{2}} d x = - \frac{11}{9 ( x - 2 )}

= \frac{7}{27} ln ∣ x + 1 ∣ - (- \frac{4}{9 ( x + 1 )}) + \frac{20}{27} ln ∣ x - 2 ∣ - \frac{11}{9 ( x - 2 )}

Simplify

= \frac{7}{27} ln ∣ x + 1 ∣ + \frac{4}{9 ( x + 1 )} + \frac{20}{27} ln ∣ x - 2 ∣ - \frac{11}{9 ( x - 2 )}

Add a constant to the solution

= \frac{7}{27} ln ∣ x + 1 ∣ + \frac{4}{9 ( x + 1 )} + \frac{20}{27} ln ∣ x - 2 ∣ - \frac{11}{9 ( x - 2 )} + C

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What is the integral of (x^3+2x-1)/((x^2-x-2)^2) ?
The integral of (x^3+2x-1)/((x^2-x-2)^2) is 7/27 ln|x+1|+4/(9(x+1))+20/27 ln|x-2|-(11)/(9(x-2))+C