Integral x/sqrt(x^2-4x)?

Hello,

let's complete the square inside the root by adding and subtracting 4:

∫ {x /√[(x² - 4x + 4) - 4]} dx =

let's write the radicand as a difference of two squares:

∫ {x /√[(x - 2)² - 2²]} dx =

let:

x - 2 = 2secθ → secθ = (x - 2)/2

x = 2secθ + 2

dx = 2tanθ secθ dθ

yielding, by trig subtitution:

∫ {x /√[(x - 2)² - 2²]} dx = ∫ {(2secθ + 2) /√[(2secθ)² - 2²]} 2tanθ secθ dθ =

∫ [(2secθ + 2) /√(2²sec²θ - 2²)] 2tanθ secθ dθ =

∫ {(2secθ + 2) /√[2²(sec²θ - 1)]} 2tanθ secθ dθ =

(applying the identity sec²θ - 1 = tan²θ)

∫ [(2secθ + 2) /√(2²tan²θ)] 2tanθ secθ dθ =

∫ [(2secθ + 2) /(2tanθ)] 2tanθ secθ dθ =

simplifying to:

∫ [(2secθ + 2) secθ dθ =

∫ (2sec²θ + 2secθ) dθ =

(splitting into two integrals and factoring constants out)

2 ∫ sec²θ dθ + 2 ∫ secθ dθ =

2tanθ + 2 ∫ secθ dθ =

le's multiply the remaining integrand by (tanθ + secθ) /(secθ +

tanθ) (= 1):

2tanθ + 2 ∫ secθ [(tanθ + secθ) /(secθ + tanθ)] dθ =

obtaining in the numerator the derivative of the denominator:

2tanθ + 2 ∫ [(tanθ secθ + sec²θ) /(secθ + tanθ)] dθ =

2tanθ + 2 ∫ d(secθ + tanθ) /(secθ + tanθ) =

2tanθ + 2 ln |secθ + tanθ| + C

let's recall that secθ = (x - 2)/2

hence:

tanθ = √(sec²θ - 1) = √{[(x - 2)/2]² - 1} =

√{[(x² - 4x + 4)/4] - 1} =

√[(x² - 4x + 4 - 4)/4] =

[√(x² - 4x)] /2

then, substituting back:

2tanθ + 2 ln |secθ + tanθ| + C = 2 {[√(x² - 4x)] /2} + 2 ln | [(x -

2)/2] + {[√(x² - 4x)] /2} | + C =

√(x² - 4x) + 2 ln | [(x - 2) + √(x² - 4x)] /2 | + C =

(by logarithm properties)

√(x² - 4x) + 2 [ln |x - 2 + √(x² - 4x)| - ln 2} + C =

√(x² - 4x) + 2 ln |x - 2 + √(x² - 4x)| - 2 ln 2 + C =

being - 2 ln 2 a constant, it can be included in C, ending with:

√(x² - 4x) + 2 ln |x - 2 + √(x² - 4x)| + C

I hope it's helpful

∫ x dx / sqrt(x^2-4x)

x^2-4x= x^2-4x+4-4 = (x-2)^2 - 4

∫ x dx / sqrt(x^2-4x) = ∫ x dx / sqrt( (x-2)^2 -4)

Let u= x-2

du = dx

∫ x dx / sqrt( (x-2)^2 -4) = ∫ (u+2) du / sqrt( u^2-4)

Let u= 2 sec t

du = 2 sec t tan t dt

u^2-4 = 4 sec^2 t - 4 = 4(sec^2 t -1) = 4 tan^2 t

sqrt(u^2-4) = 2 tan t

∫ (u+2) du / sqrt( u^2-4) = ∫ (2 sec t + 2) ( 2 sec t tan t ) dt / (2 tan t)

= ∫ (2 sec t + 2) sec t dt

= 2 ∫ sec^2 t dt + 2 ∫ sec t dt

= 2 tan t + 2 ln ( sec t + tan t)

transform t back to u

u = 2 sec t

sec t = u/2

tan t = sqrt(sec^2 t -1) = sqrt( u^2 /4 -1) = sqrt(u^2-4) / sqrt(4) = (1/2) sqrt(u^2-4)

2 tan t + 2 ln ( sec t + tan t) = 2 (1/2) sqrt(u^2-4) + 2 ln ( u / 2 + sqrt( u^2-4) / 2)

= sqrt(u^2-4) + 2 ln ( u / 2 + sqrt( u^2-4) / 2)

replace u by x-2

= sqrt( (x-2)^2 -4) + 2 ln ( (x-2) / 2 + sqrt( (x-2)^2 - 4) / 2 )

= sqrt( x^2-4x) + 2 ln ( ( x-2) / 2 + sqrt(x^2-4x) /2 )

∫ x dx / sqrt(x^2-4x) = sqrt( x^2-4x) + 2 ln ( ( x-2) / 2 + sqrt(x^2-4x) /2 ) + C