What is the integral of 8/((x^3-16x)) ?

The integral of 8/((x^3-16x)) is -8(1/16 ln|x|-1/32 ln|x+4|-1/32 ln|x-4|)+C

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integral of 8/((x^3-16x))

integral

\frac{8}{( x ^{3} - 16 x )}

- 8 (\frac{1}{16} ln ∣ x ∣ - \frac{1}{32} ln ∣ x + 4 ∣ - \frac{1}{32} ln ∣ x - 4 ∣) + C

Solution steps

\int \frac{8}{x ^{3} - 16 x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int \frac{1}{x ^{3} - 16 x} d x

Factor

x^{3} - 16 x : - (- x^{3} + 16 x)

= 8 \cdot \int \frac{1}{- ( - x ^{3} + 16 x )} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 (- \int \frac{1}{- x ^{3} + 16 x} d x)

Take the partial fraction of

\frac{1}{- x ^{3} + 16 x} : \frac{1}{16 x} - \frac{1}{32 ( x + 4 )} - \frac{1}{32 ( x - 4 )}

= 8 (- \int \frac{1}{16 x} - \frac{1}{32 ( x + 4 )} - \frac{1}{32 ( 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

= 8 (- (\int \frac{1}{16 x} d x - \int \frac{1}{32 ( x + 4 )} d x - \int \frac{1}{32 ( x - 4 )} d x))

\int \frac{1}{16 x} d x = \frac{1}{16} ln ∣ x ∣

\int \frac{1}{32 ( x + 4 )} d x = \frac{1}{32} ln ∣ x + 4 ∣

\int \frac{1}{32 ( x - 4 )} d x = \frac{1}{32} ln ∣ x - 4 ∣

= 8 (- (\frac{1}{16} ln ∣ x ∣ - \frac{1}{32} ln ∣ x + 4 ∣ - \frac{1}{32} ln ∣ x - 4 ∣))

Simplify

= - 8 (\frac{1}{16} ln ∣ x ∣ - \frac{1}{32} ln ∣ x + 4 ∣ - \frac{1}{32} ln ∣ x - 4 ∣)

Add a constant to the solution

= - 8 (\frac{1}{16} ln ∣ x ∣ - \frac{1}{32} ln ∣ x + 4 ∣ - \frac{1}{32} ln ∣ x - 4 ∣) + C

What is the integral of 8/((x^3-16x)) ?
The integral of 8/((x^3-16x)) is -8(1/16 ln|x|-1/32 ln|x+4|-1/32 ln|x-4|)+C