What is the integral of (x^4-8)/(x^3+2x^2) ?

The integral of (x^4-8)/(x^3+2x^2) is (x^2)/2-2x+2ln|x|+4/x+2ln|x+2|+C

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integral of (x^4-8)/(x^3+2x^2)

\int \frac{x ^{4} - 8}{x ^{3} + 2 x ^{2}} d x

\frac{x ^{2}}{2} - 2 x + 2 ln ∣ x ∣ + \frac{4}{x} + 2 ln ∣ x + 2 ∣ + C

Solution steps

\int \frac{x ^{4} - 8}{x ^{3} + 2 x ^{2}} d x

Take the partial fraction of

\frac{x ^{4} - 8}{x ^{3} + 2 x ^{2}} : x - 2 + \frac{2}{x} - \frac{4}{x ^{2}} + \frac{2}{x + 2}

= \int x - 2 + \frac{2}{x} - \frac{4}{x ^{2}} + \frac{2}{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 2 d x + \int \frac{2}{x} d x - \int \frac{4}{x ^{2}} d x + \int \frac{2}{x + 2} d x

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

\int 2 d x = 2 x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

\int \frac{4}{x ^{2}} d x = - \frac{4}{x}

\int \frac{2}{x + 2} d x = 2 ln ∣ x + 2 ∣

= \frac{x ^{2}}{2} - 2 x + 2 ln ∣ x ∣ - (- \frac{4}{x}) + 2 ln ∣ x + 2 ∣

Simplify

= \frac{x ^{2}}{2} - 2 x + 2 ln ∣ x ∣ + \frac{4}{x} + 2 ln ∣ x + 2 ∣

Add a constant to the solution

= \frac{x ^{2}}{2} - 2 x + 2 ln ∣ x ∣ + \frac{4}{x} + 2 ln ∣ x + 2 ∣ + C

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What is the integral of (x^4-8)/(x^3+2x^2) ?
The integral of (x^4-8)/(x^3+2x^2) is (x^2)/2-2x+2ln|x|+4/x+2ln|x+2|+C