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

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

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

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

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

Solution steps

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

Long division

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

\frac{d}{d x} (\frac{x ^{4}}{4} + \frac{8}{9} x^{3} - x^{2} + 3 x - 8)

x \to - 8 lim (\frac{∣ x - 8 ∣}{x - 8})

x \to 0 lim (\frac{x ^{3}}{3})

\frac{d}{d x} (- 5 x^{4})

x \to 3 + lim (\frac{b x - 3 b}{x - 3})

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