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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\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}))

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

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

Simplify

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

Add a constant to the solution

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

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