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

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

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

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

- 2 ln ∣ x ∣ + 2 ln x^{2} + 4 + \frac{1}{2} arctan (\frac{x}{2}) + C

Solution steps

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

Take the partial fraction of

\frac{2 x ^{2} + x - 8}{x ^{3} + 4 x} : - \frac{2}{x} + \frac{4 x + 1}{x ^{2} + 4}

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

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

Add a constant to the solution

= - 2 ln ∣ x ∣ + 2 ln x^{2} + 4 + \frac{1}{2} arctan (\frac{x}{2}) + C

\frac{\partial}{\partial y} (\frac{x ^{2} - y ^{2}}{z ^{2}})

\frac{x}{x + \frac{d}{x}}

f (x) = (2 x^{5} - 3)^{8}

x \to + (- 2.4) lim (x)

y^{^{'}} - t y^{3} = 0

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