What is the integral of (x^5)/(x^4-1) ?

The integral of (x^5)/(x^4-1) is (x^2)/2-1/4 (ln|x^2+1|-ln|x^2-1|)+C

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integral of (x^5)/(x^4-1)

\int \frac{x ^{5}}{x ^{4} - 1} d x

\frac{x ^{2}}{2} - \frac{1}{4} (ln x^{2} + 1 - ln x^{2} - 1) + C

Solution steps

\int \frac{x ^{5}}{x ^{4} - 1} d x

Long division

\frac{x ^{5}}{x ^{4} - 1} : x + \frac{x}{x ^{4} - 1}

= \int x + \frac{x}{x ^{4} - 1} 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 x d x + \int \frac{x}{x ^{4} - 1} d x

\int x d x = \frac{x ^{2}}{2}

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

= \frac{x ^{2}}{2} - \frac{1}{4} (ln x^{2} + 1 - ln x^{2} - 1)

Add a constant to the solution

= \frac{x ^{2}}{2} - \frac{1}{4} (ln x^{2} + 1 - ln x^{2} - 1) + C

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What is the integral of (x^5)/(x^4-1) ?
The integral of (x^5)/(x^4-1) is (x^2)/2-1/4 (ln|x^2+1|-ln|x^2-1|)+C