What is the integral of x/(sqrt(x^2-4x)) ?

The integral of x/(sqrt(x^2-4x)) is sqrt(x^2-4x)+2ln(1/2 |sqrt(x^2-4x)+x-2|)+C

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integral of x/(sqrt(x^2-4x))

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

x^{2} - 4 x + 2 ln (\frac{1}{2} x^{2} - 4 x + x - 2) + C

Solution steps

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

Complete the square

x^{2} - 4 x : (x - 2)^{2} - 4

= \int \frac{x}{( x - 2 ) ^{2} - 4} d x

Apply u-substitution

: \int \frac{u + 2}{u ^{2} - 4} d u

= \int \frac{u + 2}{u ^{2} - 4} d u

Expand

\frac{u + 2}{u ^{2} - 4} : \frac{u}{u ^{2} - 4} + \frac{2}{u ^{2} - 4}

= \int \frac{u}{u ^{2} - 4} + \frac{2}{u ^{2} - 4} 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} - 4} d u + \int \frac{2}{u ^{2} - 4} d u

\int \frac{u}{u ^{2} - 4} d u = u^{2} - 4

\int \frac{2}{u ^{2} - 4} d u = 2 ln (\frac{1}{2} u^{2} - 4 + u)

= u^{2} - 4 + 2 ln (\frac{1}{2} u^{2} - 4 + u)

Substitute back

u = x - 2

= (x - 2)^{2} - 4 + 2 ln (\frac{1}{2} (x - 2)^{2} - 4 + x - 2)

Simplify

(x - 2)^{2} - 4 + 2 ln (\frac{1}{2} (x - 2)^{2} - 4 + x - 2) : x^{2} - 4 x + 2 ln (\frac{1}{2} x^{2} - 4 x + x - 2)

= x^{2} - 4 x + 2 ln (\frac{1}{2} x^{2} - 4 x + x - 2)

Add a constant to the solution

= x^{2} - 4 x + 2 ln (\frac{1}{2} x^{2} - 4 x + x - 2) + C

- \frac{1}{s ( s ^{2} + 1 )}

y^{^{''}} + 4 y = 8 sin (2 x)

\frac{3 s - 1}{s ^{2} + 2 s + 5}

sin^{2} (x) + cos^{2} (x)

x \to 0.5 lim (\frac{sin ( x )}{x})

What is the integral of x/(sqrt(x^2-4x)) ?
The integral of x/(sqrt(x^2-4x)) is sqrt(x^2-4x)+2ln(1/2 |sqrt(x^2-4x)+x-2|)+C