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

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

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

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

x^{2} - 2 x - ln x^{2} - 2 x + x - 1 + C

Solution steps

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

Complete the square

x^{2} - 2 x : (x - 1)^{2} - 1

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

Apply u-substitution

: \int \frac{u ^{2} - 1}{u + 1} d u

= \int \frac{u ^{2} - 1}{u + 1} d u

\frac{u ^{2} - 1}{u + 1} = \frac{( u ^{2} - 1 ) ( u - 1 )}{( u ) ^{2} - ( 1 ) ^{2}}

= \int \frac{u ^{2} - 1 ( u - 1 )}{u ^{2} - 1 ^{2}} d u

Simplify

\frac{u ^{2} - 1 ( u - 1 )}{u ^{2} - 1 ^{2}} : \frac{u - 1}{u ^{2} - 1}

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

Expand

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

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

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

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

\int \frac{1}{u ^{2} - 1} d u = ln u^{2} - 1 + u

= u^{2} - 1 - ln u^{2} - 1 + u

Substitute back

u = x - 1

= (x - 1)^{2} - 1 - ln (x - 1)^{2} - 1 + x - 1

Simplify

(x - 1)^{2} - 1 - ln (x - 1)^{2} - 1 + x - 1 : x^{2} - 2 x - ln x^{2} - 2 x + x - 1

= x^{2} - 2 x - ln x^{2} - 2 x + x - 1

Add a constant to the solution

= x^{2} - 2 x - ln x^{2} - 2 x + x - 1 + C

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