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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

= \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

x \to 0 lim (x^{2} sin (x))

\int - 5 sin (5 t) d t

y^{3}

\int_{0}^{x} sin^{2} (t) d t

\frac{d x}{d y}, y^{2} - x = 0

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