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

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

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

Expand

\frac{x - 4}{x ^{2} - 4 x} : \frac{x}{x ^{2} - 4 x} - \frac{4}{x ^{2} - 4 x}

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

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

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

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

Simplify

= 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

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