What is the integral of (x^5-2x^3-2x)/(x^2+1) ?

The integral of (x^5-2x^3-2x)/(x^2+1) is (x^4)/4-(x^2)/2-x^2+1/2 ln|x^2+1|-1+C

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integral of (x^5-2x^3-2x)/(x^2+1)

\int \frac{x ^{5} - 2 x ^{3} - 2 x}{x ^{2} + 1} d x

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

Solution steps

\int \frac{x ^{5} - 2 x ^{3} - 2 x}{x ^{2} + 1} d x

Expand

\frac{x ^{5} - 2 x ^{3} - 2 x}{x ^{2} + 1} : \frac{x ^{5}}{x ^{2} + 1} - \frac{2 x ^{3}}{x ^{2} + 1} - \frac{2 x}{x ^{2} + 1}

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{x ^{5}}{x ^{2} + 1} d x - \int \frac{2 x ^{3}}{x ^{2} + 1} d x - \int \frac{2 x}{x ^{2} + 1} d x

\int \frac{x ^{5}}{x ^{2} + 1} d x = \frac{x ^{4}}{4} - \frac{x ^{2}}{2} + \frac{1}{2} ln x^{2} + 1

\int \frac{2 x ^{3}}{x ^{2} + 1} d x = x^{2} + 1 - ln x^{2} + 1

\int \frac{2 x}{x ^{2} + 1} d x = ln x^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of (x^5-2x^3-2x)/(x^2+1) ?
The integral of (x^5-2x^3-2x)/(x^2+1) is (x^4)/4-(x^2)/2-x^2+1/2 ln|x^2+1|-1+C