What is the integral of 1/((x+3)(x+2)) ?

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

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integral of 1/((x+3)(x+2))

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

- ln ∣ x + 3 ∣ + ln ∣ x + 2 ∣ + C

Solution steps

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

Take the partial fraction of

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

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

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

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

= - ln ∣ x + 3 ∣ + ln ∣ x + 2 ∣

Add a constant to the solution

= - ln ∣ x + 3 ∣ + ln ∣ x + 2 ∣ + C

(4 x^{3} + 4 x y - 1) d x - (1 - 2 x^{2} - 2 y) d y = 0

f (x) = e^{x} x

\int t e^{- 10 t} d t

\int \frac{cos ( x )}{2 - cos ( x )} d x

\frac{d}{d y} (\frac{x ^{2}}{y})

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