What is the integral of (-x^6-x^4+x^2+6)/(x^2+1) ?

The integral of (-x^6-x^4+x^2+6)/(x^2+1) is -(x^5)/5+x+5arctan(x)+C

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integral of (-x^6-x^4+x^2+6)/(x^2+1)

\int \frac{- x ^{6} - x ^{4} + x ^{2} + 6}{x ^{2} + 1} d x

- \frac{x ^{5}}{5} + x + 5 arctan (x) + C

Solution steps

\int \frac{- x ^{6} - x ^{4} + x ^{2} + 6}{x ^{2} + 1} d x

Expand

\frac{- x ^{6} - x ^{4} + x ^{2} + 6}{x ^{2} + 1} : - \frac{x ^{6}}{x ^{2} + 1} - \frac{x ^{4}}{x ^{2} + 1} + \frac{x ^{2}}{x ^{2} + 1} + \frac{6}{x ^{2} + 1}

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 ^{6}}{x ^{2} + 1} d x - \int \frac{x ^{4}}{x ^{2} + 1} d x + \int \frac{x ^{2}}{x ^{2} + 1} d x + \int \frac{6}{x ^{2} + 1} d x

\int \frac{x ^{6}}{x ^{2} + 1} d x = \frac{x ^{5}}{5} - \frac{x ^{3}}{3} + x - arctan (x)

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

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

\int \frac{6}{x ^{2} + 1} d x = 6 arctan (x)

= - (\frac{x ^{5}}{5} - \frac{x ^{3}}{3} + x - arctan (x)) - (\frac{x ^{3}}{3} - x + arctan (x)) - arctan (x) + x + 6 arctan (x)

Simplify

- (\frac{x ^{5}}{5} - \frac{x ^{3}}{3} + x - arctan (x)) - (\frac{x ^{3}}{3} - x + arctan (x)) - arctan (x) + x + 6 arctan (x) : - \frac{x ^{5}}{5} + x + 5 arctan (x)

= - \frac{x ^{5}}{5} + x + 5 arctan (x)

Add a constant to the solution

= - \frac{x ^{5}}{5} + x + 5 arctan (x) + C

\frac{\partial}{\partial x} (x^{5} + 4 z^{2} e^{y - x})

\int x^{3} e^{- \frac{x}{3}} d x

x^{4}, x

\frac{\partial}{\partial y} (ln (y))

\frac{\frac{E}{L}}{( s ^{2} + \frac{1}{C L} )}

What is the integral of (-x^6-x^4+x^2+6)/(x^2+1) ?
The integral of (-x^6-x^4+x^2+6)/(x^2+1) is -(x^5)/5+x+5arctan(x)+C