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|-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 x^{2} + 9 - 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 x^{2} + 9 - arctan (\frac{x}{3}))

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{2}{7} sec (x) tan (x) d x

\frac{d y}{d t} = \frac{6 t ^{5}}{1 + t ^{6}} y + 7 (1 + t^{6})^{2}

t^{10}

sin^{3} (x co) s^{5} x

\int 5 arctan (\frac{1}{x}) d x

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|-arctan(x/3)+C