What is the integral of 1/((x-6)(x-3)) ?

The integral of 1/((x-6)(x-3)) is 1/3 ln|x-6|-1/3 ln|x-3|+C

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integral of 1/((x-6)(x-3))

\int \frac{1}{( x - 6 ) ( x - 3 )} d x

\frac{1}{3} ln ∣ x - 6 ∣ - \frac{1}{3} ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{1}{( x - 6 ) ( x - 3 )} d x

Take the partial fraction of

\frac{1}{( x - 6 ) ( x - 3 )} : \frac{1}{3 ( x - 6 )} - \frac{1}{3 ( x - 3 )}

= \int \frac{1}{3 ( x - 6 )} - \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

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

\int \frac{1}{3 ( x - 6 )} d x = \frac{1}{3} ln ∣ x - 6 ∣

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

= \frac{1}{3} ln ∣ x - 6 ∣ - \frac{1}{3} ln ∣ x - 3 ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ x - 6 ∣ - \frac{1}{3} ln ∣ x - 3 ∣ + C

\frac{\partial}{\partial x} (\frac{93}{2 + x ^{2} + y ^{2}})

x \to 1 lim (\frac{x}{∣ x ∣})

y = (ln (x))^{l n (x)}

\frac{\partial}{\partial x} (\frac{z}{x + y})

\frac{d y}{d x} = \frac{( 2 x + 5 )}{( x ^{2} + 2 x )}, y (1) = 1

What is the integral of 1/((x-6)(x-3)) ?
The integral of 1/((x-6)(x-3)) is 1/3 ln|x-6|-1/3 ln|x-3|+C