What is the integral of (x-3)/(x^2-7x-18) ?

The integral of (x-3)/(x^2-7x-18) is 5/11 ln|x+2|+6/11 ln|x-9|+C

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integral of (x-3)/(x^2-7x-18)

\int \frac{x - 3}{x ^{2} - 7 x - 18} d x

\frac{5}{11} ln ∣ x + 2 ∣ + \frac{6}{11} ln ∣ x - 9 ∣ + C

Solution steps

\int \frac{x - 3}{x ^{2} - 7 x - 18} d x

Take the partial fraction of

\frac{x - 3}{x ^{2} - 7 x - 18} : \frac{5}{11 ( x + 2 )} + \frac{6}{11 ( x - 9 )}

= \int \frac{5}{11 ( x + 2 )} + \frac{6}{11 ( x - 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

= \int \frac{5}{11 ( x + 2 )} d x + \int \frac{6}{11 ( x - 9 )} d x

\int \frac{5}{11 ( x + 2 )} d x = \frac{5}{11} ln ∣ x + 2 ∣

\int \frac{6}{11 ( x - 9 )} d x = \frac{6}{11} ln ∣ x - 9 ∣

= \frac{5}{11} ln ∣ x + 2 ∣ + \frac{6}{11} ln ∣ x - 9 ∣

Add a constant to the solution

= \frac{5}{11} ln ∣ x + 2 ∣ + \frac{6}{11} ln ∣ x - 9 ∣ + C

\frac{\partial}{\partial x} (- 6 e^{6 y} sin (6 x))

\frac{d y}{d x} = \frac{( x - 2 )}{y - 2}

\int ln (x + 2) d x

5 y^{^{'}} - y = e^{- t} + 3 t

\int \frac{( 9 sin ^{3} ( x ) )}{cos ( x )} d x

What is the integral of (x-3)/(x^2-7x-18) ?
The integral of (x-3)/(x^2-7x-18) is 5/11 ln|x+2|+6/11 ln|x-9|+C