What is the integral of (x^4)/(x^3-2x^2-7x-4) ?

The integral of (x^4)/(x^3-2x^2-7x-4) is (x^2)/2+2x+19/25 ln|x+1|+1/(5(x+1))+256/25 ln|x-4|+C

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integral of (x^4)/(x^3-2x^2-7x-4)

\int \frac{x ^{4}}{x ^{3} - 2 x ^{2} - 7 x - 4} d x

\frac{x ^{2}}{2} + 2 x + \frac{19}{25} ln ∣ x + 1 ∣ + \frac{1}{5 ( x + 1 )} + \frac{256}{25} ln ∣ x - 4 ∣ + C

Solution steps

\int \frac{x ^{4}}{x ^{3} - 2 x ^{2} - 7 x - 4} d x

Take the partial fraction of

\frac{x ^{4}}{x ^{3} - 2 x ^{2} - 7 x - 4} : x + 2 + \frac{19}{25 ( x + 1 )} - \frac{1}{5 ( x + 1 ) ^{2}} + \frac{256}{25 ( x - 4 )}

= \int x + 2 + \frac{19}{25 ( x + 1 )} - \frac{1}{5 ( x + 1 ) ^{2}} + \frac{256}{25 ( x - 4 )} 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 2 d x + \int \frac{19}{25 ( x + 1 )} d x - \int \frac{1}{5 ( x + 1 ) ^{2}} d x + \int \frac{256}{25 ( x - 4 )} d x

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

\int 2 d x = 2 x

\int \frac{19}{25 ( x + 1 )} d x = \frac{19}{25} ln ∣ x + 1 ∣

\int \frac{1}{5 ( x + 1 ) ^{2}} d x = - \frac{1}{5 ( x + 1 )}

\int \frac{256}{25 ( x - 4 )} d x = \frac{256}{25} ln ∣ x - 4 ∣

= \frac{x ^{2}}{2} + 2 x + \frac{19}{25} ln ∣ x + 1 ∣ - (- \frac{1}{5 ( x + 1 )}) + \frac{256}{25} ln ∣ x - 4 ∣

Simplify

= \frac{x ^{2}}{2} + 2 x + \frac{19}{25} ln ∣ x + 1 ∣ + \frac{1}{5 ( x + 1 )} + \frac{256}{25} ln ∣ x - 4 ∣

Add a constant to the solution

= \frac{x ^{2}}{2} + 2 x + \frac{19}{25} ln ∣ x + 1 ∣ + \frac{1}{5 ( x + 1 )} + \frac{256}{25} ln ∣ x - 4 ∣ + C

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What is the integral of (x^4)/(x^3-2x^2-7x-4) ?
The integral of (x^4)/(x^3-2x^2-7x-4) is (x^2)/2+2x+19/25 ln|x+1|+1/(5(x+1))+256/25 ln|x-4|+C