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

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

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

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

- \frac{x ^{3}}{3} + 2 x - 2 arctan (x) + C

Solution steps

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

Expand

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

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

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

\int \frac{x ^{4}}{1 + x ^{2}} d x = \frac{x ^{3}}{3} - x + arctan (x)

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

Simplify

- arctan (x) + x - (\frac{x ^{3}}{3} - x + arctan (x)) : - \frac{x ^{3}}{3} + 2 x - 2 arctan (x)

= - \frac{x ^{3}}{3} + 2 x - 2 arctan (x)

Add a constant to the solution

= - \frac{x ^{3}}{3} + 2 x - 2 arctan (x) + C

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