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

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

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

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

ln ∣ x ∣ - 5 ln ∣ x + 2 ∣ + 5 ln ∣ x - 2 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2} + 20 x - 4}{x ^{3} - 4 x} : \frac{1}{x} - \frac{5}{x + 2} + \frac{5}{x - 2}

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

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{5}{x + 2} d x = 5 ln ∣ x + 2 ∣

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

= ln ∣ x ∣ - 5 ln ∣ x + 2 ∣ + 5 ln ∣ x - 2 ∣

Add a constant to the solution

= ln ∣ x ∣ - 5 ln ∣ x + 2 ∣ + 5 ln ∣ x - 2 ∣ + C

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

(12 x^{2} - 20 y^{2}) d x + (- 20 x y + 36 x^{- 1} y^{3}) d y = 0

\int e^{x} d x

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

y = 5 x^{3} + 4 x^{2} - 6, (1, 3)

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