What is the integral of 1/(x^4-16) ?

The integral of 1/(x^4-16) is -1/16 arctan(x/2)-1/32 ln|x+2|+1/32 ln|x-2|+C

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integral of 1/(x^4-16)

\int \frac{1}{x ^{4} - 16} d x

- \frac{1}{16} arctan (\frac{x}{2}) - \frac{1}{32} ln ∣ x + 2 ∣ + \frac{1}{32} ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{1}{x ^{4} - 16} d x

Take the partial fraction of

\frac{1}{x ^{4} - 16} : - \frac{1}{8 ( x ^{2} + 4 )} - \frac{1}{32 ( x + 2 )} + \frac{1}{32 ( x - 2 )}

= \int - \frac{1}{8 ( x ^{2} + 4 )} - \frac{1}{32 ( x + 2 )} + \frac{1}{32 ( 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}{8 ( x ^{2} + 4 )} d x - \int \frac{1}{32 ( x + 2 )} d x + \int \frac{1}{32 ( x - 2 )} d x

\int \frac{1}{8 ( x ^{2} + 4 )} d x = \frac{1}{16} arctan (\frac{x}{2})

\int \frac{1}{32 ( x + 2 )} d x = \frac{1}{32} ln ∣ x + 2 ∣

\int \frac{1}{32 ( x - 2 )} d x = \frac{1}{32} ln ∣ x - 2 ∣

= - \frac{1}{16} arctan (\frac{x}{2}) - \frac{1}{32} ln ∣ x + 2 ∣ + \frac{1}{32} ln ∣ x - 2 ∣

Add a constant to the solution

= - \frac{1}{16} arctan (\frac{x}{2}) - \frac{1}{32} ln ∣ x + 2 ∣ + \frac{1}{32} ln ∣ x - 2 ∣ + C

\int_{0}^{1} 5 e^{5 x} d x

\frac{d}{d x} (\frac{9 x}{x})

x \to 0 + lim (x^{x^{8}})

\int 36 x^{3} (3 x^{4} + 3)^{5} d x

y \to 0 lim (\frac{0}{y})

What is the integral of 1/(x^4-16) ?
The integral of 1/(x^4-16) is -1/16 arctan(x/2)-1/32 ln|x+2|+1/32 ln|x-2|+C