What is the integral of (x^4)/((x-1)^2(x+2)) ?

The integral of (x^4)/((x-1)^2(x+2)) is (x^2)/2+11/9 ln|x-1|-1/(3(x-1))+16/9 ln|x+2|+C

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integral of (x^4)/((x-1)^2(x+2))

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

\frac{x ^{2}}{2} + \frac{11}{9} ln ∣ x - 1 ∣ - \frac{1}{3 ( x - 1 )} + \frac{16}{9} ln ∣ x + 2 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x ^{4}}{( x - 1 ) ^{2} ( x + 2 )} : x + \frac{11}{9 ( x - 1 )} + \frac{1}{3 ( x - 1 ) ^{2}} + \frac{16}{9 ( x + 2 )}

= \int x + \frac{11}{9 ( x - 1 )} + \frac{1}{3 ( x - 1 ) ^{2}} + \frac{16}{9 ( x + 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 x d x + \int \frac{11}{9 ( x - 1 )} d x + \int \frac{1}{3 ( x - 1 ) ^{2}} d x + \int \frac{16}{9 ( x + 2 )} d x

\int x d x = \frac{x ^{2}}{2}

\int \frac{11}{9 ( x - 1 )} d x = \frac{11}{9} ln ∣ x - 1 ∣

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

\int \frac{16}{9 ( x + 2 )} d x = \frac{16}{9} ln ∣ x + 2 ∣

= \frac{x ^{2}}{2} + \frac{11}{9} ln ∣ x - 1 ∣ - \frac{1}{3 ( x - 1 )} + \frac{16}{9} ln ∣ x + 2 ∣

Add a constant to the solution

= \frac{x ^{2}}{2} + \frac{11}{9} ln ∣ x - 1 ∣ - \frac{1}{3 ( x - 1 )} + \frac{16}{9} ln ∣ x + 2 ∣ + C

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What is the integral of (x^4)/((x-1)^2(x+2)) ?
The integral of (x^4)/((x-1)^2(x+2)) is (x^2)/2+11/9 ln|x-1|-1/(3(x-1))+16/9 ln|x+2|+C