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

\int arcsin (t) d t

\frac{\partial}{\partial x} (2 x^{2} - 5 y^{2})

\int_{- 2}^{6} 4 x - x^{2} + 12 d x

\int_{5}^{7} [\frac{k}{2} x^{2}] d x

y^{^{''}} + 5 y^{^{'}} = e^{- t}, y (0) = 0, y^{^{'}} (0) = 1

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