What is the integral of (x^3-4x^2-4x+20)/(x^2-5) ?

The integral of (x^3-4x^2-4x+20)/(x^2-5) is (x^2)/2-4x+1/2 ln|x^2-5|+C

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integral of (x^3-4x^2-4x+20)/(x^2-5)

\int \frac{x ^{3} - 4 x ^{2} - 4 x + 20}{x ^{2} - 5} d x

\frac{x ^{2}}{2} - 4 x + \frac{1}{2} ln x^{2} - 5 + C

Solution steps

\int \frac{x ^{3} - 4 x ^{2} - 4 x + 20}{x ^{2} - 5} d x

Long division

\frac{x ^{3} - 4 x ^{2} - 4 x + 20}{x ^{2} - 5} : x - 4 + \frac{x}{x ^{2} - 5}

= \int x - 4 + \frac{x}{x ^{2} - 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 d x - \int 4 d x + \int \frac{x}{x ^{2} - 5} d x

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

\int 4 d x = 4 x

\int \frac{x}{x ^{2} - 5} d x = \frac{1}{2} ln x^{2} - 5

= \frac{x ^{2}}{2} - 4 x + \frac{1}{2} ln x^{2} - 5

Add a constant to the solution

= \frac{x ^{2}}{2} - 4 x + \frac{1}{2} ln x^{2} - 5 + C

f (x) = 8 e^{- x} + e^{3 x}

\frac{s + 1}{s ^{2}}

f (x) = 1 x^{3}

\frac{d}{d x} (8 sin (\frac{2 π}{3}))

\int \frac{2 x}{3 x - 5} d x

What is the integral of (x^3-4x^2-4x+20)/(x^2-5) ?
The integral of (x^3-4x^2-4x+20)/(x^2-5) is (x^2)/2-4x+1/2 ln|x^2-5|+C