What is the integral of 1/((x+2)(x+4)(x+6)) ?

The integral of 1/((x+2)(x+4)(x+6)) is 1/8 ln|x+2|-1/4 ln|x+4|+1/8 ln|x+6|+C

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integral of 1/((x+2)(x+4)(x+6))

\int \frac{1}{( x + 2 ) ( x + 4 ) ( x + 6 )} d x

\frac{1}{8} ln ∣ x + 2 ∣ - \frac{1}{4} ln ∣ x + 4 ∣ + \frac{1}{8} ln ∣ x + 6 ∣ + C

Solution steps

\int \frac{1}{( x + 2 ) ( x + 4 ) ( x + 6 )} d x

Take the partial fraction of

\frac{1}{( x + 2 ) ( x + 4 ) ( x + 6 )} : \frac{1}{8 ( x + 2 )} - \frac{1}{4 ( x + 4 )} + \frac{1}{8 ( x + 6 )}

= \int \frac{1}{8 ( x + 2 )} - \frac{1}{4 ( x + 4 )} + \frac{1}{8 ( 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

= \int \frac{1}{8 ( x + 2 )} d x - \int \frac{1}{4 ( x + 4 )} d x + \int \frac{1}{8 ( x + 6 )} d x

\int \frac{1}{8 ( x + 2 )} d x = \frac{1}{8} ln ∣ x + 2 ∣

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

\int \frac{1}{8 ( x + 6 )} d x = \frac{1}{8} ln ∣ x + 6 ∣

= \frac{1}{8} ln ∣ x + 2 ∣ - \frac{1}{4} ln ∣ x + 4 ∣ + \frac{1}{8} ln ∣ x + 6 ∣

Add a constant to the solution

= \frac{1}{8} ln ∣ x + 2 ∣ - \frac{1}{4} ln ∣ x + 4 ∣ + \frac{1}{8} ln ∣ x + 6 ∣ + C

\frac{d}{d x} (lo g_{e} (x^{x}))

cos^{2} (θ)

\frac{d y}{d x} = y (950 - y)

\int \frac{7}{x + 7 + x} d x

\frac{d}{d x} (\frac{x ^{3} y - x y ^{3}}{x ^{2} + y ^{2}})

What is the integral of 1/((x+2)(x+4)(x+6)) ?
The integral of 1/((x+2)(x+4)(x+6)) is 1/8 ln|x+2|-1/4 ln|x+4|+1/8 ln|x+6|+C