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

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

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

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

2 (- ln ∣ x - 2 ∣ + ln ∣ x - 1 ∣) + C

Solution steps

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

Take the partial fraction of

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

\int \frac{x ^{3} + x ^{2} sin ( x ) - 2}{x ^{2}} d x

\frac{2}{x}

\int \frac{x ^{2} + 3}{x ^{2} - 4} d x

\int \frac{x ^{4}}{x ^{4} + 2 x ^{2} + 1} d x

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

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