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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Take the partial fraction of

\frac{1}{x ( x + 3 )} : \frac{1}{3 x} - \frac{1}{3 ( x + 3 )}

= 9 \cdot \int \frac{1}{3 x} - \frac{1}{3 ( 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

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

\int \frac{1}{3 x} d x = \frac{1}{3} ln ∣ x ∣

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

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

Add a constant to the solution

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

\int ln (y) d y

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\frac{d y}{d x} = \frac{( x - y )}{x + 2 y}, y (0) = 1

f^{^{''}} (x) = x^{\frac{1}{14}}

\int_{1}^{2} 3^{- θ} d θ

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