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

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

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

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

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

Solution steps

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

Long division

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

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

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

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

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

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

Add a constant to the solution

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

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