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

The integral of 1/(sqrt((x-1)(x-2))) is ln|2sqrt(x^2-3x+2)+2x-3|+C

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

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

ln 2 x^{2} - 3 x + 2 + 2 x - 3 + C

Solution steps

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

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

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

\int \frac{1 - ( x - 1 ) ( x - 2 )}{( x - 1 ) ( x - 2 )} d x = ln 2 x^{2} - 3 x + 2 + 2 x - 3 - x

\int 1 d x = x

= ln 2 x^{2} - 3 x + 2 + 2 x - 3 - x + x

Simplify

= ln 2 x^{2} - 3 x + 2 + 2 x - 3

Add a constant to the solution

= ln 2 x^{2} - 3 x + 2 + 2 x - 3 + C

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What is the integral of 1/(sqrt((x-1)(x-2))) ?
The integral of 1/(sqrt((x-1)(x-2))) is ln|2sqrt(x^2-3x+2)+2x-3|+C