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

The integral of 1/((x-6)(x-3)(x+1)) is 1/21 ln|x-6|-1/12 ln|x-3|+1/28 ln|x+1|+C

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

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

\frac{1}{21} ln ∣ x - 6 ∣ - \frac{1}{12} ln ∣ x - 3 ∣ + \frac{1}{28} ln ∣ x + 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{( x - 6 ) ( x - 3 ) ( x + 1 )} : \frac{1}{21 ( x - 6 )} - \frac{1}{12 ( x - 3 )} + \frac{1}{28 ( x + 1 )}

= \int \frac{1}{21 ( x - 6 )} - \frac{1}{12 ( x - 3 )} + \frac{1}{28 ( x + 1 )} 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}{21 ( x - 6 )} d x - \int \frac{1}{12 ( x - 3 )} d x + \int \frac{1}{28 ( x + 1 )} d x

\int \frac{1}{21 ( x - 6 )} d x = \frac{1}{21} ln ∣ x - 6 ∣

\int \frac{1}{12 ( x - 3 )} d x = \frac{1}{12} ln ∣ x - 3 ∣

\int \frac{1}{28 ( x + 1 )} d x = \frac{1}{28} ln ∣ x + 1 ∣

= \frac{1}{21} ln ∣ x - 6 ∣ - \frac{1}{12} ln ∣ x - 3 ∣ + \frac{1}{28} ln ∣ x + 1 ∣

Add a constant to the solution

= \frac{1}{21} ln ∣ x - 6 ∣ - \frac{1}{12} ln ∣ x - 3 ∣ + \frac{1}{28} ln ∣ x + 1 ∣ + C

y^{^{'}} = 100 x - x y

\frac{\partial}{\partial x} (- 9 x^{2} y^{2} - cos (y))

x^{3} + 8

\frac{1}{18 x ^{2}}, x, \frac{x}{8}

\frac{d ^{2}}{d x ^{2}} (e^{- 6 x})

What is the integral of 1/((x-6)(x-3)(x+1)) ?
The integral of 1/((x-6)(x-3)(x+1)) is 1/21 ln|x-6|-1/12 ln|x-3|+1/28 ln|x+1|+C