What is the integral of 1/((x+2)(x+5)) ?

The integral of 1/((x+2)(x+5)) is 1/3 ln|x+2|-1/3 ln|x+5|+C

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integral of 1/((x+2)(x+5))

\int \frac{1}{( x + 2 ) ( x + 5 )} d x

\frac{1}{3} ln ∣ x + 2 ∣ - \frac{1}{3} ln ∣ x + 5 ∣ + C

Solution steps

\int \frac{1}{( x + 2 ) ( x + 5 )} d x

Take the partial fraction of

\frac{1}{( x + 2 ) ( x + 5 )} : \frac{1}{3 ( x + 2 )} - \frac{1}{3 ( x + 5 )}

= \int \frac{1}{3 ( x + 2 )} - \frac{1}{3 ( x + 5 )} 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{1}{3 ( x + 5 )} d x

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

\int \frac{1}{3 ( x + 5 )} d x = \frac{1}{3} ln ∣ x + 5 ∣

= \frac{1}{3} ln ∣ x + 2 ∣ - \frac{1}{3} ln ∣ x + 5 ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ x + 2 ∣ - \frac{1}{3} ln ∣ x + 5 ∣ + C

\frac{\partial}{\partial x} (\frac{x}{y} + sin (2 x + 3 y))

\int_{0}^{6} \frac{1}{u ^{2} - 36} d u

x \to 2 lim (\frac{∣ x - 2 ∣}{x - 2})

\int b cos (x) d x

\frac{d y}{d x} = \frac{y}{x} + 9 x + 4

What is the integral of 1/((x+2)(x+5)) ?
The integral of 1/((x+2)(x+5)) is 1/3 ln|x+2|-1/3 ln|x+5|+C