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

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

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

\int \frac{x - 1}{x ^{2} - 2 x} d x

x^{2} - 2 x + C

Solution steps

\int \frac{x - 1}{x ^{2} - 2 x} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

= \int \frac{1}{2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= \frac{1}{2} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = x^{2} - 2 x

= \frac{1}{2} \cdot 2 (x^{2} - 2 x)^{\frac{1}{2}}

Simplify

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

= x^{2} - 2 x

Add a constant to the solution

= x^{2} - 2 x + C

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\int \frac{4 x}{16 - x ^{2}} d x

\int 3 sec^{2} (3 x) tan (3 x) d x

\int \frac{( arcsin ( ln ( x ) ) )}{x} d x

\frac{1}{x ^{4}}

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