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

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

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

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

- \frac{x ^{2}}{2} + 3 ln x^{2} + 1 + C

Solution steps

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

Long division

\frac{x ( 5 - x ^{2} )}{x ^{2} + 1} : - x + \frac{6 x}{x ^{2} + 1}

= \int - x + \frac{6 x}{x ^{2} + 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{6 x}{x ^{2} + 1} d x

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

\int \frac{6 x}{x ^{2} + 1} d x = 3 ln x^{2} + 1

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

Add a constant to the solution

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

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