What is the integral of (x^4)/((x^2+1)) ?

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

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integral of (x^4)/((x^2+1))

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

\frac{x ^{3}}{3} - x + arctan (x) + C

Solution steps

\int \frac{x ^{4}}{x ^{2} + 1} d x

Long division

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

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

\int x^{2} d x = \frac{x ^{3}}{3}

\int 1 d x = x

\int \frac{1}{x ^{2} + 1} d x = arctan (x)

= \frac{x ^{3}}{3} - x + arctan (x)

Add a constant to the solution

= \frac{x ^{3}}{3} - x + arctan (x) + C

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