What is the integral of ((x^3-6x-20))/(x+5) ?

The integral of ((x^3-6x-20))/(x+5) is (x^3)/3-(5x^2)/2+19x-115ln|x+5|+C

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integral of ((x^3-6x-20))/(x+5)

\int \frac{( x ^{3} - 6 x - 20 )}{x + 5} d x

\frac{x ^{3}}{3} - \frac{5 x ^{2}}{2} + 19 x - 115 ln ∣ x + 5 ∣ + C

Solution steps

\int \frac{x ^{3} - 6 x - 20}{x + 5} d x

Long division

\frac{x ^{3} - 6 x - 20}{x + 5} : x^{2} - 5 x + 19 - \frac{115}{x + 5}

= \int x^{2} - 5 x + 19 - \frac{115}{x + 5} 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^{2} d x - \int 5 x d x + \int 19 d x - \int \frac{115}{x + 5} d x

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

\int 5 x d x = \frac{5 x ^{2}}{2}

\int 19 d x = 19 x

\int \frac{115}{x + 5} d x = 115 ln ∣ x + 5 ∣

= \frac{x ^{3}}{3} - \frac{5 x ^{2}}{2} + 19 x - 115 ln ∣ x + 5 ∣

Add a constant to the solution

= \frac{x ^{3}}{3} - \frac{5 x ^{2}}{2} + 19 x - 115 ln ∣ x + 5 ∣ + C

y = (x^{4}) + (\frac{3}{( x ^{2} )}), at x = - 1

y^{^{'}} + 2 y = t

e^{- a x^{2}}, x = 0

(x + 1) \cdot \frac{d y}{d x} + y = ln (x)

\frac{d}{d x} (e^{- 2 x} + e^{3 x})

What is the integral of ((x^3-6x-20))/(x+5) ?
The integral of ((x^3-6x-20))/(x+5) is (x^3)/3-(5x^2)/2+19x-115ln|x+5|+C