What is the integral of (x^2+16x-4)/(x^3-4x) ?

The integral of (x^2+16x-4)/(x^3-4x) is ln|x|-4ln|x+2|+4ln|x-2|+C

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integral of (x^2+16x-4)/(x^3-4x)

\int \frac{x ^{2} + 16 x - 4}{x ^{3} - 4 x} d x

ln ∣ x ∣ - 4 ln ∣ x + 2 ∣ + 4 ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{x ^{2} + 16 x - 4}{x ^{3} - 4 x} d x

Take the partial fraction of

\frac{x ^{2} + 16 x - 4}{x ^{3} - 4 x} : \frac{1}{x} - \frac{4}{x + 2} + \frac{4}{x - 2}

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

= ln ∣ x ∣ - 4 ln ∣ x + 2 ∣ + 4 ln ∣ x - 2 ∣

Add a constant to the solution

= ln ∣ x ∣ - 4 ln ∣ x + 2 ∣ + 4 ln ∣ x - 2 ∣ + C

y = \frac{1}{3 + 2 x}, (2, \frac{1}{7})

7 x - 6 y x^{2} + 1 \frac{d y}{d x} = 0

\frac{d y}{d x} = (e^{4 x + 3 y})

(sin (π) x)^{^{'}}

\frac{d}{d x} (ln (sin (x)))

What is the integral of (x^2+16x-4)/(x^3-4x) ?
The integral of (x^2+16x-4)/(x^3-4x) is ln|x|-4ln|x+2|+4ln|x-2|+C