Question

Find or evaluate the integral. (Complete the square, if necessary.) $$ \int \frac{1}{(x-1) \sqrt{x^{2}-2 x}} d x $$

Short answer

The value of the integral is given by \( \int \frac{1}{(x-1) \sqrt{x^{2}-2 x}} dx = \arcsin (x - 1) + C \).

Step-by-step solution

Complete the Square

To complete the square, group the x term. The square can be expressed as \( x^2 - 2x = (x-1)^2 - 1 \). Therefore, the integral expression now reads \( \int \frac{1}{(x-1) \sqrt{{(x-1)^2 - 1}}} dx \).

Substitution

Next, perform the substitution \( u = x - 1 \) to simplify. The differential du is determined to be \( dx \) and the new integral becomes \( \int \frac{1}{u \sqrt{u^2 - 1}} du \).

Identified Integral

The integral has now been transformed into an identified integral in form of \( \int \frac{1}{a \sqrt{a^2 - x^2}} dx = \arcsin (\frac{x}{a}) \). In this case, \( a = 1 \) and \( x = u \), hence the integral reduces to \( \int \frac{1}{u \sqrt{1 - {u^2}}} du = \arcsin{u} \).

Substitute back and Simplify

Substitute the value of u back into the equation \( u = x - 1 \). After substituting, the integral becomes \( \arcsin (x - 1) + C \), where C is the constant of integration.