What is the integral of (x^4-x^3+2x^2-2x-8)/(x^2+4) ?

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

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integral of (x^4-x^3+2x^2-2x-8)/(x^2+4)

\int \frac{x ^{4} - x ^{3} + 2 x ^{2} - 2 x - 8}{x ^{2} + 4} d x

\frac{x ^{3}}{3} - \frac{1}{2} x^{2} - 2 x + ln x^{2} + 4 - 2 + C

Solution steps

\int \frac{x ^{4} - x ^{3} + 2 x ^{2} - 2 x - 8}{x ^{2} + 4} d x

Expand

\frac{x ^{4} - x ^{3} + 2 x ^{2} - 2 x - 8}{x ^{2} + 4} : \frac{x ^{4}}{x ^{2} + 4} - \frac{x ^{3}}{x ^{2} + 4} + \frac{2 x ^{2}}{x ^{2} + 4} - \frac{2 x}{x ^{2} + 4} - \frac{8}{x ^{2} + 4}

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 ^{2} + 4} d x - \int \frac{x ^{3}}{x ^{2} + 4} d x + \int \frac{2 x ^{2}}{x ^{2} + 4} d x - \int \frac{2 x}{x ^{2} + 4} d x - \int \frac{8}{x ^{2} + 4} d x

\int \frac{x ^{4}}{x ^{2} + 4} d x = 8 (\frac{x ^{3}}{24} - \frac{x}{2} + arctan (\frac{x}{2}))

\int \frac{x ^{3}}{x ^{2} + 4} d x = \frac{1}{2} (x^{2} + 4 - 4 ln x^{2} + 4)

\int \frac{2 x ^{2}}{x ^{2} + 4} d x = 4 (- arctan (\frac{x}{2}) + \frac{x}{2})

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

\int \frac{8}{x ^{2} + 4} d x = 4 arctan (\frac{x}{2})

= 8 (\frac{x ^{3}}{24} - \frac{x}{2} + arctan (\frac{x}{2})) - \frac{1}{2} (x^{2} + 4 - 4 ln x^{2} + 4) + 4 (- arctan (\frac{x}{2}) + \frac{x}{2}) - ln x^{2} + 4 - 4 arctan (\frac{x}{2})

Simplify

8 (\frac{x ^{3}}{24} - \frac{x}{2} + arctan (\frac{x}{2})) - \frac{1}{2} (x^{2} + 4 - 4 ln x^{2} + 4) + 4 (- arctan (\frac{x}{2}) + \frac{x}{2}) - ln x^{2} + 4 - 4 arctan (\frac{x}{2}) : \frac{x ^{3}}{3} - \frac{1}{2} x^{2} - 2 x + ln x^{2} + 4 - 2

= \frac{x ^{3}}{3} - \frac{1}{2} x^{2} - 2 x + ln x^{2} + 4 - 2

Add a constant to the solution

= \frac{x ^{3}}{3} - \frac{1}{2} x^{2} - 2 x + ln x^{2} + 4 - 2 + C

x \to 4 lim (\frac{x}{x ^{2} - 3 x})

y^{^{'}} = (x + 1) y^{2}

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

\frac{d}{d x} (\frac{d y}{d x}) = 4 x

f (x) = - 3 e^{- 2 x} + 3

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