What is the integral of 1/((x-2)(x-4)(x+1)) ?

The integral of 1/((x-2)(x-4)(x+1)) is -1/6 ln|x-2|+1/10 ln|x-4|+1/15 ln|x+1|+C

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integral of 1/((x-2)(x-4)(x+1))

\int \frac{1}{( x - 2 ) ( x - 4 ) ( x + 1 )} d x

- \frac{1}{6} ln ∣ x - 2 ∣ + \frac{1}{10} ln ∣ x - 4 ∣ + \frac{1}{15} ln ∣ x + 1 ∣ + C

Solution steps

\int \frac{1}{( x - 2 ) ( x - 4 ) ( x + 1 )} d x

Take the partial fraction of

\frac{1}{( x - 2 ) ( x - 4 ) ( x + 1 )} : - \frac{1}{6 ( x - 2 )} + \frac{1}{10 ( x - 4 )} + \frac{1}{15 ( x + 1 )}

= \int - \frac{1}{6 ( x - 2 )} + \frac{1}{10 ( x - 4 )} + \frac{1}{15 ( 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}{6 ( x - 2 )} d x + \int \frac{1}{10 ( x - 4 )} d x + \int \frac{1}{15 ( x + 1 )} d x

\int \frac{1}{6 ( x - 2 )} d x = \frac{1}{6} ln ∣ x - 2 ∣

\int \frac{1}{10 ( x - 4 )} d x = \frac{1}{10} ln ∣ x - 4 ∣

\int \frac{1}{15 ( x + 1 )} d x = \frac{1}{15} ln ∣ x + 1 ∣

= - \frac{1}{6} ln ∣ x - 2 ∣ + \frac{1}{10} ln ∣ x - 4 ∣ + \frac{1}{15} ln ∣ x + 1 ∣

Add a constant to the solution

= - \frac{1}{6} ln ∣ x - 2 ∣ + \frac{1}{10} ln ∣ x - 4 ∣ + \frac{1}{15} ln ∣ x + 1 ∣ + C

\frac{d}{d x} (a x^{4} - 2 a x^{3} + 2 b x^{2} \cdot ln (\frac{x}{2}))

x + 2

x = - y^{2} + 4 y, [1, 4]

x \to 0 lim (sin^{2} (\frac{π}{x}))

\frac{\partial}{\partial x} ((9 x + 2 y)^{10})

What is the integral of 1/((x-2)(x-4)(x+1)) ?
The integral of 1/((x-2)(x-4)(x+1)) is -1/6 ln|x-2|+1/10 ln|x-4|+1/15 ln|x+1|+C