What is the integral of (x^3+9)/(x^2+9) ?

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

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integral of (x^3+9)/(x^2+9)

\int \frac{x ^{3} + 9}{x ^{2} + 9} d x

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

Solution steps

\int \frac{x ^{3} + 9}{x ^{2} + 9} d x

Apply Integral Substitution

= \int \frac{3 ( 3 u ^{3} + 1 )}{u ^{2} + 1} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int \frac{3 u ^{3} + 1}{u ^{2} + 1} d u

Expand

\frac{3 u ^{3} + 1}{u ^{2} + 1} : \frac{3 u ^{3}}{u ^{2} + 1} + \frac{1}{u ^{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

= 3 (\int \frac{3 u ^{3}}{u ^{2} + 1} d u + \int \frac{1}{u ^{2} + 1} d u)

\int \frac{3 u ^{3}}{u ^{2} + 1} d u = \frac{3}{2} (u^{2} + 1 - ln u^{2} + 1)

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

= 3 (\frac{3}{2} (u^{2} + 1 - ln u^{2} + 1) + arctan (u))

Substitute back

u = \frac{x}{3}

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

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{x ^{2} + 1}{3 - x ^{2}})

\frac{d}{d x} ((x - π) sin (x))

\frac{1}{3 x + 4}

x \to 4 lim (- \frac{3 x}{7} + \frac{8}{7})

x \to 9 - lim (81 - x^{2})

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