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

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

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

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

ln ∣ x - 2 ∣ - ln ∣ x - 1 ∣ + C

Solution steps

\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}

= \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

= \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 ∣

= ln ∣ x - 2 ∣ - ln ∣ x - 1 ∣

Add a constant to the solution

= ln ∣ x - 2 ∣ - ln ∣ x - 1 ∣ + C

\frac{7}{1 - t}

\frac{\partial}{\partial x} (t e^{- t})

\frac{\partial}{\partial x} (2 x^{2} - 8 x + y^{2} - 8 y + 7)

\frac{d}{d x} (\frac{ln ( x ^{3} - y + z )}{x y + z})

\int - 2 x + 2 d x

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