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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

\int \frac{1}{2 ( x + 4 )} d x = \frac{1}{2} ln ∣ x + 4 ∣

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

Add a constant to the solution

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

(1 + x y) d x + x^{2} d y = 0

\int - sin (x) tan (x) d x

in v erse l a pl a ce \frac{1}{( s + 2 ) ^{2} + 1}

d er i v a t i v e \frac{8 - x e ^{x}}{x + e ^{x}}

\int \frac{x + x ^{3}}{x} - e^{x} d x

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