What is the integral of (x^2+60x+60)/(x^3-4x) ?

The integral of (x^2+60x+60)/(x^3-4x) is -15ln|x|-7ln|x+2|+23ln|x-2|+C

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integral of (x^2+60x+60)/(x^3-4x)

\int \frac{x ^{2} + 60 x + 60}{x ^{3} - 4 x} d x

- 15 ln ∣ x ∣ - 7 ln ∣ x + 2 ∣ + 23 ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{x ^{2} + 60 x + 60}{x ^{3} - 4 x} d x

Take the partial fraction of

\frac{x ^{2} + 60 x + 60}{x ^{3} - 4 x} : - \frac{15}{x} - \frac{7}{x + 2} + \frac{23}{x - 2}

= \int - \frac{15}{x} - \frac{7}{x + 2} + \frac{23}{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{15}{x} d x - \int \frac{7}{x + 2} d x + \int \frac{23}{x - 2} d x

\int \frac{15}{x} d x = 15 ln ∣ x ∣

\int \frac{7}{x + 2} d x = 7 ln ∣ x + 2 ∣

\int \frac{23}{x - 2} d x = 23 ln ∣ x - 2 ∣

= - 15 ln ∣ x ∣ - 7 ln ∣ x + 2 ∣ + 23 ln ∣ x - 2 ∣

Add a constant to the solution

= - 15 ln ∣ x ∣ - 7 ln ∣ x + 2 ∣ + 23 ln ∣ x - 2 ∣ + C

n = 1 \sum \infty (- 0.3)^{n - 1}

x \to 3 + lim (\frac{5 x}{x ^{2} - 9})

\frac{d}{d x} (ln (cosh (2 x)))

\frac{d}{d x} (sin (x) + \frac{7}{4} cot (x))

2 y^{^{''}} + y^{^{'}} - 10 y = 0

What is the integral of (x^2+60x+60)/(x^3-4x) ?
The integral of (x^2+60x+60)/(x^3-4x) is -15ln|x|-7ln|x+2|+23ln|x-2|+C