What is the integral of 1/((x+4)(x+8)) ?

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

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integral of 1/((x+4)(x+8))

\int \frac{1}{( x + 4 ) ( x + 8 )} d x

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

Solution steps

\int \frac{1}{( x + 4 ) ( x + 8 )} d x

Take the partial fraction of

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

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

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

\int \frac{1}{4 ( x + 8 )} d x = \frac{1}{4} ln ∣ x + 8 ∣

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

Add a constant to the solution

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

(3, - 2), (7, 0)

x \to 1 lim (\frac{x - 2}{x ^{2} - 1})

(3.12) (5.6)

y^{^{''}} + y = 7 e^{- 5 x}, y (0) = 0, y^{^{'}} (0) = 0

\frac{\partial}{\partial x} (4 y x^{3} - 15 y^{2} x^{4})

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