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

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

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integral of 2/((x-2)(x-1))

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

2 (ln ∣ x - 2 ∣ - ln ∣ x - 1 ∣) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

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

= 2 \cdot \int \frac{1}{x - 2} - \frac{1}{x - 1} 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

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

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

\int \frac{1}{x - 1} d x = ln ∣ x - 1 ∣

= 2 (ln ∣ x - 2 ∣ - ln ∣ x - 1 ∣)

Add a constant to the solution

= 2 (ln ∣ x - 2 ∣ - ln ∣ x - 1 ∣) + C

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