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

The integral of 9/(x^2+3x-18) is ln|x-3|-ln|x+6|+C

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

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

ln ∣ x - 3 ∣ - ln ∣ x + 6 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{x ^{2} + 3 x - 18} : \frac{1}{9 ( x - 3 )} - \frac{1}{9 ( x + 6 )}

= 9 \cdot \int \frac{1}{9 ( x - 3 )} - \frac{1}{9 ( x + 6 )} 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}{9 ( x - 3 )} d x - \int \frac{1}{9 ( x + 6 )} d x)

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

\int \frac{1}{9 ( x + 6 )} d x = \frac{1}{9} ln ∣ x + 6 ∣

= 9 (\frac{1}{9} ln ∣ x - 3 ∣ - \frac{1}{9} ln ∣ x + 6 ∣)

Simplify

9 (\frac{1}{9} ln ∣ x - 3 ∣ - \frac{1}{9} ln ∣ x + 6 ∣) : ln ∣ x - 3 ∣ - ln ∣ x + 6 ∣

= ln ∣ x - 3 ∣ - ln ∣ x + 6 ∣

Add a constant to the solution

= ln ∣ x - 3 ∣ - ln ∣ x + 6 ∣ + C

\int_{- 3}^{1} ∣ 3 x + 3 ∣ d x

x^{2} y^{^{'}} = 2 y^{2} - x^{2}

\int sinh^{2} (2 x) d x

b e^{θ}

\frac{\partial}{\partial x} (x^{2} + 2 y^{4} + 8 x y)

What is the integral of 9/(x^2+3x-18) ?
The integral of 9/(x^2+3x-18) is ln|x-3|-ln|x+6|+C