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

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

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integral of 1/(16-x^4)

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

Take the partial fraction of

\frac{1}{16 - x ^{4}} : \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 sin (2 x) e^{3 x} d x

\int 24 x^{5} arccot (2 x^{3}) d x

\int \frac{1}{x ^{2} ( x ^{2} + 4 )} d x

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