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|-1/4 ln|x+2|+1/4 ln|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 ∣ - \frac{1}{4} ln ∣ x + 2 ∣ + \frac{1}{4} ln ∣ 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{1}{4 ( x + 2 )} + \frac{1}{4 ( x - 2 )}

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

\int \frac{1}{4 ( x - 2 )} d x = \frac{1}{4} ln ∣ x - 2 ∣

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

Add a constant to the solution

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

(2 x - 1) y^{^{'}} + y^{2} = 0

y^{^{''}} + 9 y^{^{'}} + 14 y = 0, y (0) = 1, y^{^{'}} (0) = 0

x^{2} + 5

\frac{d ^{2} y}{d t ^{2}} + 5 \frac{d y}{d t} = e^{- t}

5 y^{^{''}} + 35 y = 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|-1/4 ln|x+2|+1/4 ln|x-2|+C