What is the integral of 1/(16x-x^3) ?

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

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integral of 1/(16x-x^3)

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

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

Solution steps

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

Take the partial fraction of

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

= \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

= \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 ∣

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

Add a constant to the solution

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

\int \frac{1}{x + 4} e^{x + 4} d x

\int arctan (\frac{y}{x}) d x

\int \frac{ln ( x ^{2} )}{x ^{4}} d x

\int \frac{( x ^{3} + 1 )}{x ^{4} + 4 x} d x

2

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