What is the integral of (x^4+1)/(x^3+3x^2+2x) ?

The integral of (x^4+1)/(x^3+3x^2+2x) is (x^2)/2-3x+1/2 ln|x|+17/2 ln|x+2|-2ln|x+1|+C

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integral of (x^4+1)/(x^3+3x^2+2x)

\int \frac{x ^{4} + 1}{x ^{3} + 3 x ^{2} + 2 x} d x

\frac{x ^{2}}{2} - 3 x + \frac{1}{2} ln ∣ x ∣ + \frac{17}{2} ln ∣ x + 2 ∣ - 2 ln ∣ x + 1 ∣ + C

Solution steps

\int \frac{x ^{4} + 1}{x ^{3} + 3 x ^{2} + 2 x} d x

Expand

\frac{x ^{4} + 1}{x ^{3} + 3 x ^{2} + 2 x} : \frac{x ^{4}}{x ^{3} + 3 x ^{2} + 2 x} + \frac{1}{x ^{3} + 3 x ^{2} + 2 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{x ^{4}}{x ^{3} + 3 x ^{2} + 2 x} d x + \int \frac{1}{x ^{3} + 3 x ^{2} + 2 x} d x

\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 ∣

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

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

Simplify

\frac{x ^{2}}{2} - 3 x - ln ∣ x + 1 ∣ + 8 ln ∣ x + 2 ∣ + \frac{1}{2} ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x + 2 ∣ : \frac{x ^{2}}{2} - 3 x + \frac{1}{2} ln ∣ x ∣ + \frac{17}{2} ln ∣ x + 2 ∣ - 2 ln ∣ x + 1 ∣

= \frac{x ^{2}}{2} - 3 x + \frac{1}{2} ln ∣ x ∣ + \frac{17}{2} ln ∣ x + 2 ∣ - 2 ln ∣ x + 1 ∣

Add a constant to the solution

= \frac{x ^{2}}{2} - 3 x + \frac{1}{2} ln ∣ x ∣ + \frac{17}{2} ln ∣ x + 2 ∣ - 2 ln ∣ x + 1 ∣ + C

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What is the integral of (x^4+1)/(x^3+3x^2+2x) ?
The integral of (x^4+1)/(x^3+3x^2+2x) is (x^2)/2-3x+1/2 ln|x|+17/2 ln|x+2|-2ln|x+1|+C