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

The integral of (x^4)/(x^3+3x^2+2x) is (x^2)/2-3x-ln|x+1|+8ln|x+2|+C

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

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

\frac{x ^{2}}{2} - 3 x - ln ∣ x + 1 ∣ + 8 ln ∣ x + 2 ∣ + C

Solution steps

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

Simplify

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

= \int \frac{x ^{3}}{x ^{2} + 3 x + 2} d x

Long division

\frac{x ^{3}}{x ^{2} + 3 x + 2} : x - 3 + \frac{7 x + 6}{x ^{2} + 3 x + 2}

= \int x - 3 + \frac{7 x + 6}{x ^{2} + 3 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 3 d x + \int \frac{7 x + 6}{x ^{2} + 3 x + 2} d x

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

\int 3 d x = 3 x

\int \frac{7 x + 6}{x ^{2} + 3 x + 2} d x = - ln ∣ x + 1 ∣ + 8 ln ∣ x + 2 ∣

= \frac{x ^{2}}{2} - 3 x - ln ∣ x + 1 ∣ + 8 ln ∣ x + 2 ∣

Add a constant to the solution

= \frac{x ^{2}}{2} - 3 x - ln ∣ x + 1 ∣ + 8 ln ∣ x + 2 ∣ + C

What is the integral of (x^4)/(x^3+3x^2+2x) ?
The integral of (x^4)/(x^3+3x^2+2x) is (x^2)/2-3x-ln|x+1|+8ln|x+2|+C