What is the integral of 3/(x^4-16) ?

The integral of 3/(x^4-16) is 3(-1/16 arctan(x/2)-1/32 ln|x+2|+1/32 ln|x-2|)+C

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integral of 3/(x^4-16)

\int \frac{3}{x ^{4} - 16} d x

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

Solution steps

\int \frac{3}{x ^{4} - 16} d x

Take the constant out:

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

= 3 \cdot \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 )}

= 3 \cdot \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

= 3 (- \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 ∣

= 3 (- \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

= 3 (- \frac{1}{16} arctan (\frac{x}{2}) - \frac{1}{32} ln ∣ x + 2 ∣ + \frac{1}{32} ln ∣ x - 2 ∣) + C

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What is the integral of 3/(x^4-16) ?
The integral of 3/(x^4-16) is 3(-1/16 arctan(x/2)-1/32 ln|x+2|+1/32 ln|x-2|)+C