What is the integral of (x^3+14)/(x^2+3x+2) ?

The integral of (x^3+14)/(x^2+3x+2) is (x^2)/2-3x+13ln|x+1|-6ln|x+2|+C

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integral of (x^3+14)/(x^2+3x+2)

\int \frac{x ^{3} + 14}{x ^{2} + 3 x + 2} d x

\frac{x ^{2}}{2} - 3 x + 13 ln ∣ x + 1 ∣ - 6 ln ∣ x + 2 ∣ + C

Solution steps

\int \frac{x ^{3} + 14}{x ^{2} + 3 x + 2} d x

Long division

\frac{x ^{3} + 14}{x ^{2} + 3 x + 2} : x - 3 + \frac{7 x + 20}{x ^{2} + 3 x + 2}

= \int x - 3 + \frac{7 x + 20}{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 + 20}{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 + 20}{x ^{2} + 3 x + 2} d x = 13 ln ∣ x + 1 ∣ - 6 ln ∣ x + 2 ∣

= \frac{x ^{2}}{2} - 3 x + 13 ln ∣ x + 1 ∣ - 6 ln ∣ x + 2 ∣

Add a constant to the solution

= \frac{x ^{2}}{2} - 3 x + 13 ln ∣ x + 1 ∣ - 6 ln ∣ x + 2 ∣ + C

cos (2 t)

\frac{d y}{d x} = y^{2} + 1, y (4) = 0

x \to 1 lim (\frac{cos ( x )}{x})

n = 1 \sum \infty 5^{- n}

\frac{\partial}{\partial x} (arcsin (\frac{l}{x}))

What is the integral of (x^3+14)/(x^2+3x+2) ?
The integral of (x^3+14)/(x^2+3x+2) is (x^2)/2-3x+13ln|x+1|-6ln|x+2|+C