What is the integral of (18)/((x+3)(x^2+9)) ?

The integral of (18)/((x+3)(x^2+9)) is ln|x+3|-1/2 ln|(x^2)/9+1|+arctan(x/3)+C

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integral of (18)/((x+3)(x^2+9))

\int \frac{18}{( x + 3 ) ( x ^{2} + 9 )} d x

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

Solution steps

\int \frac{18}{( x + 3 ) ( x ^{2} + 9 )} d x

Take the constant out:

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

= 18 \cdot \int \frac{1}{( x + 3 ) ( x ^{2} + 9 )} d x

Take the partial fraction of

\frac{1}{( x + 3 ) ( x ^{2} + 9 )} : \frac{1}{18 ( x + 3 )} + \frac{- x + 3}{18 ( x ^{2} + 9 )}

= 18 \cdot \int \frac{1}{18 ( x + 3 )} + \frac{- x + 3}{18 ( x ^{2} + 9 )} 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

= 18 (\int \frac{1}{18 ( x + 3 )} d x + \int \frac{- x + 3}{18 ( x ^{2} + 9 )} d x)

\int \frac{1}{18 ( x + 3 )} d x = \frac{1}{18} ln ∣ x + 3 ∣

\int \frac{- x + 3}{18 ( x ^{2} + 9 )} d x = \frac{1}{18} (- \frac{1}{2} ln \frac{x ^{2}}{9} + 1 + arctan (\frac{x}{3}))

= 18 (\frac{1}{18} ln ∣ x + 3 ∣ + \frac{1}{18} (- \frac{1}{2} ln \frac{x ^{2}}{9} + 1 + arctan (\frac{x}{3})))

Simplify

18 (\frac{1}{18} ln ∣ x + 3 ∣ + \frac{1}{18} (- \frac{1}{2} ln \frac{x ^{2}}{9} + 1 + arctan (\frac{x}{3}))) : ln ∣ x + 3 ∣ - \frac{1}{2} ln \frac{x ^{2}}{9} + 1 + arctan (\frac{x}{3})

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

Add a constant to the solution

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

x \to 2 lim (- 6)

\frac{d}{d x} (π x)

\int \frac{1}{9 x ^{2} - 18 x + 72} d x

\int - 4 ln (2 x + 1) d x

(y^{- 1} + y^{- 2}) (2 y^{- 3} - 6 y^{- 4})

What is the integral of (18)/((x+3)(x^2+9)) ?
The integral of (18)/((x+3)(x^2+9)) is ln|x+3|-1/2 ln|(x^2)/9+1|+arctan(x/3)+C