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

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

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

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

- ln ∣ x - 1 ∣ + ln ∣ x - 2 ∣ + C

Solution steps

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

Take the partial fraction of

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

\frac{1}{cos ^{2} ( x )}

\frac{d x}{d t} = x

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

x \to \infty lim \frac{( ( lo g _{2} ( x ) ) ^{2} )}{x}

x y y^{^{'}} = x^{6} + 3 y^{2}

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