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

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

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

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

- ln ∣ x - 1 ∣ + ln ∣ x - 2 ∣ + C

Solution steps

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

Take the partial fraction of

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

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

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

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

= - ln ∣ x - 1 ∣ + ln ∣ x - 2 ∣

Add a constant to the solution

= - ln ∣ x - 1 ∣ + ln ∣ x - 2 ∣ + C

x \to - 3 lim (\frac{5}{x + 3})

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

\int_{1}^{4} (\frac{x}{2} - \frac{5}{x}) d x

x \to \frac{7}{8} lim (x^{4})

\frac{d}{d x} (\frac{e ^{x - 2}}{x})

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