What is the integral of (x^4+9x^2+x+2)/(x^2+9) ?

The integral of (x^4+9x^2+x+2)/(x^2+9) is (x^3)/3+1/2 ln|x^2+9|+2/3 arctan(x/3)+C

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integral of (x^4+9x^2+x+2)/(x^2+9)

\int \frac{x ^{4} + 9 x ^{2} + x + 2}{x ^{2} + 9} d x

\frac{x ^{3}}{3} + \frac{1}{2} ln x^{2} + 9 + \frac{2}{3} arctan (\frac{x}{3}) + C

Solution steps

\int \frac{x ^{4} + 9 x ^{2} + x + 2}{x ^{2} + 9} d x

Expand

\frac{x ^{4} + 9 x ^{2} + x + 2}{x ^{2} + 9} : \frac{x ^{4}}{x ^{2} + 9} + \frac{9 x ^{2}}{x ^{2} + 9} + \frac{x}{x ^{2} + 9} + \frac{2}{x ^{2} + 9}

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 ^{4}}{x ^{2} + 9} d x + \int \frac{9 x ^{2}}{x ^{2} + 9} d x + \int \frac{x}{x ^{2} + 9} d x + \int \frac{2}{x ^{2} + 9} d x

\int \frac{x ^{4}}{x ^{2} + 9} d x = 27 (\frac{x ^{3}}{81} - \frac{x}{3} + arctan (\frac{x}{3}))

\int \frac{9 x ^{2}}{x ^{2} + 9} d x = 27 (- arctan (\frac{x}{3}) + \frac{x}{3})

\int \frac{x}{x ^{2} + 9} d x = \frac{1}{2} ln x^{2} + 9

\int \frac{2}{x ^{2} + 9} d x = \frac{2}{3} arctan (\frac{x}{3})

= 27 (\frac{x ^{3}}{81} - \frac{x}{3} + arctan (\frac{x}{3})) + 27 (- arctan (\frac{x}{3}) + \frac{x}{3}) + \frac{1}{2} ln x^{2} + 9 + \frac{2}{3} arctan (\frac{x}{3})

Simplify

27 (\frac{x ^{3}}{81} - \frac{x}{3} + arctan (\frac{x}{3})) + 27 (- arctan (\frac{x}{3}) + \frac{x}{3}) + \frac{1}{2} ln x^{2} + 9 + \frac{2}{3} arctan (\frac{x}{3}) : \frac{x ^{3}}{3} + \frac{1}{2} ln x^{2} + 9 + \frac{2}{3} arctan (\frac{x}{3})

= \frac{x ^{3}}{3} + \frac{1}{2} ln x^{2} + 9 + \frac{2}{3} arctan (\frac{x}{3})

Add a constant to the solution

= \frac{x ^{3}}{3} + \frac{1}{2} ln x^{2} + 9 + \frac{2}{3} arctan (\frac{x}{3}) + C

x \to - 2 lim (\frac{2 - x}{2 + x})

2 x^{2} (3 x^{4} - 2)

\frac{\partial}{\partial x} (6 x sin (x^{2} y))

y = (x - 2)^{\frac{2}{3}}, (3, 1)

\int e^{- \frac{1}{3} x} d x

What is the integral of (x^4+9x^2+x+2)/(x^2+9) ?
The integral of (x^4+9x^2+x+2)/(x^2+9) is (x^3)/3+1/2 ln|x^2+9|+2/3 arctan(x/3)+C