What is the integral of x/(sqrt(x^2-8x)) ?

The integral of x/(sqrt(x^2-8x)) is sqrt(x^2-8x)+4ln(1/4 |sqrt(x^2-8x)+x-4|)+C

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integral of x/(sqrt(x^2-8x))

\int \frac{x}{x ^{2} - 8 x} d x

x^{2} - 8 x + 4 ln (\frac{1}{4} x^{2} - 8 x + x - 4) + C

Solution steps

\int \frac{x}{x ^{2} - 8 x} d x

Complete the square

x^{2} - 8 x : (x - 4)^{2} - 16

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

Apply u-substitution

: \int \frac{u + 4}{u ^{2} - 16} d u

= \int \frac{u + 4}{u ^{2} - 16} d u

Expand

\frac{u + 4}{u ^{2} - 16} : \frac{u}{u ^{2} - 16} + \frac{4}{u ^{2} - 16}

= \int \frac{u}{u ^{2} - 16} + \frac{4}{u ^{2} - 16} 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} - 16} d u + \int \frac{4}{u ^{2} - 16} d u

\int \frac{u}{u ^{2} - 16} d u = u^{2} - 16

\int \frac{4}{u ^{2} - 16} d u = 4 ln (\frac{1}{4} u^{2} - 16 + u)

= u^{2} - 16 + 4 ln (\frac{1}{4} u^{2} - 16 + u)

Substitute back

u = x - 4

= (x - 4)^{2} - 16 + 4 ln (\frac{1}{4} (x - 4)^{2} - 16 + x - 4)

Simplify

(x - 4)^{2} - 16 + 4 ln (\frac{1}{4} (x - 4)^{2} - 16 + x - 4) : x^{2} - 8 x + 4 ln (\frac{1}{4} x^{2} - 8 x + x - 4)

= x^{2} - 8 x + 4 ln (\frac{1}{4} x^{2} - 8 x + x - 4)

Add a constant to the solution

= x^{2} - 8 x + 4 ln (\frac{1}{4} x^{2} - 8 x + x - 4) + C

\frac{d}{d x} (5 x sin (x))

\frac{d}{d x} (3 x^{2} - 2 x + 4)

\frac{\partial}{\partial y} (y (1 - x^{2}))

x \to 1 lim (\frac{2 x + 3}{x + 1})

What is the integral of x/(sqrt(x^2-8x)) ?
The integral of x/(sqrt(x^2-8x)) is sqrt(x^2-8x)+4ln(1/4 |sqrt(x^2-8x)+x-4|)+C