What is the integral of (x^2)/(x^4-16) ?

The integral of (x^2)/(x^4-16) is 1/4 arctan(x/2)-1/8 ln|x+2|+1/8 ln|x-2|+C

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integral of (x^2)/(x^4-16)

\int \frac{x ^{2}}{x ^{4} - 16} d x

\frac{1}{4} arctan (\frac{x}{2}) - \frac{1}{8} ln ∣ x + 2 ∣ + \frac{1}{8} ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{x ^{2}}{x ^{4} - 16} d x

Take the partial fraction of

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

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

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

\int \frac{1}{8 ( x + 2 )} d x = \frac{1}{8} ln ∣ x + 2 ∣

\int \frac{1}{8 ( x - 2 )} d x = \frac{1}{8} ln ∣ x - 2 ∣

= \frac{1}{4} arctan (\frac{x}{2}) - \frac{1}{8} ln ∣ x + 2 ∣ + \frac{1}{8} ln ∣ x - 2 ∣

Add a constant to the solution

= \frac{1}{4} arctan (\frac{x}{2}) - \frac{1}{8} ln ∣ x + 2 ∣ + \frac{1}{8} ln ∣ x - 2 ∣ + C

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What is the integral of (x^2)/(x^4-16) ?
The integral of (x^2)/(x^4-16) is 1/4 arctan(x/2)-1/8 ln|x+2|+1/8 ln|x-2|+C