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

The integral of x/((x+2)(x-4)) is 1/3 ln|x+2|+2/3 ln|x-4|+C

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

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

\frac{1}{3} ln ∣ x + 2 ∣ + \frac{2}{3} ln ∣ x - 4 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x}{( x + 2 ) ( x - 4 )} : \frac{1}{3 ( x + 2 )} + \frac{2}{3 ( x - 4 )}

= \int \frac{1}{3 ( x + 2 )} + \frac{2}{3 ( x - 4 )} 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 + 2 )} d x + \int \frac{2}{3 ( x - 4 )} d x

\int \frac{1}{3 ( x + 2 )} d x = \frac{1}{3} ln ∣ x + 2 ∣

\int \frac{2}{3 ( x - 4 )} d x = \frac{2}{3} ln ∣ x - 4 ∣

= \frac{1}{3} ln ∣ x + 2 ∣ + \frac{2}{3} ln ∣ x - 4 ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ x + 2 ∣ + \frac{2}{3} ln ∣ x - 4 ∣ + C

\int sec (2 x) \cdot tan (2 x) d x

\int \frac{2 x - 3}{x ^{2} + 4} d x

n = 1 \sum \infty \frac{2 n}{n ^{2}}

f (x) = 7 arctan (7 x)

y = - 8 x ln (6 x + 5)

What is the integral of x/((x+2)(x-4)) ?
The integral of x/((x+2)(x-4)) is 1/3 ln|x+2|+2/3 ln|x-4|+C