What is the integral of (x^5-x)/(x^2+1) ?

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

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integral of (x^5-x)/(x^2+1)

\int \frac{x ^{5} - x}{x ^{2} + 1} d x

\frac{x ^{4}}{4} - \frac{x ^{2}}{2} + C

Solution steps

\int \frac{x ^{5} - x}{x ^{2} + 1} d x

Simplify

\frac{x ^{5} - x}{x ^{2} + 1} : x (x + 1) (x - 1)

= \int x (x + 1) (x - 1) d x

Expand

x (x + 1) (x - 1) : x^{3} - x

= \int x^{3} - x 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^{3} d x - \int x d x

\int x^{3} d x = \frac{x ^{4}}{4}

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

= \frac{x ^{4}}{4} - \frac{x ^{2}}{2}

Add a constant to the solution

= \frac{x ^{4}}{4} - \frac{x ^{2}}{2} + C

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