What is the integral of (x^2+48x-16)/(x^3-16x) ?

The integral of (x^2+48x-16)/(x^3-16x) is ln|x|-6ln|x+4|+6ln|x-4|+C

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integral of (x^2+48x-16)/(x^3-16x)

\int \frac{x ^{2} + 48 x - 16}{x ^{3} - 16 x} d x

ln ∣ x ∣ - 6 ln ∣ x + 4 ∣ + 6 ln ∣ x - 4 ∣ + C

Solution steps

\int \frac{x ^{2} + 48 x - 16}{x ^{3} - 16 x} d x

Take the partial fraction of

\frac{x ^{2} + 48 x - 16}{x ^{3} - 16 x} : \frac{1}{x} - \frac{6}{x + 4} + \frac{6}{x - 4}

= \int \frac{1}{x} - \frac{6}{x + 4} + \frac{6}{x - 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{1}{x} d x - \int \frac{6}{x + 4} d x + \int \frac{6}{x - 4} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{6}{x + 4} d x = 6 ln ∣ x + 4 ∣

\int \frac{6}{x - 4} d x = 6 ln ∣ x - 4 ∣

= ln ∣ x ∣ - 6 ln ∣ x + 4 ∣ + 6 ln ∣ x - 4 ∣

Add a constant to the solution

= ln ∣ x ∣ - 6 ln ∣ x + 4 ∣ + 6 ln ∣ x - 4 ∣ + C

\frac{d}{d x} (2^{x} + 1)

\frac{d y}{d t} = - \frac{y}{t ^{2}}

f (x) = 6 x - x^{2}, at x = 4

\frac{d}{d x} ((x - 5) sin (x))

n = 0 \sum \infty \frac{1}{1. 5 ^{n}}

What is the integral of (x^2+48x-16)/(x^3-16x) ?
The integral of (x^2+48x-16)/(x^3-16x) is ln|x|-6ln|x+4|+6ln|x-4|+C