What is the integral of (4(x-3))/(sqrt(x^2-6x)) ?

The integral of (4(x-3))/(sqrt(x^2-6x)) is 4sqrt(x^2-6x)+C

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integral of (4(x-3))/(sqrt(x^2-6x))

\int \frac{4 ( x - 3 )}{x ^{2} - 6 x} d x

4 x^{2} - 6 x + C

Solution steps

\int \frac{4 ( x - 3 )}{x ^{2} - 6 x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int \frac{x - 3}{x ^{2} - 6 x} d x

Expand

\frac{x - 3}{x ^{2} - 6 x} : \frac{x}{x ^{2} - 6 x} - \frac{3}{x ^{2} - 6 x}

= 4 \cdot \int \frac{x}{x ^{2} - 6 x} - \frac{3}{x ^{2} - 6 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

= 4 (\int \frac{x}{x ^{2} - 6 x} d x - \int \frac{3}{x ^{2} - 6 x} d x)

\int \frac{x}{x ^{2} - 6 x} d x = x^{2} - 6 x + 3 ln (\frac{1}{3} x^{2} - 6 x + x - 3)

\int \frac{3}{x ^{2} - 6 x} d x = 3 ln (\frac{1}{3} x^{2} - 6 x + x - 3)

= 4 (x^{2} - 6 x + 3 ln (\frac{1}{3} x^{2} - 6 x + x - 3) - 3 ln (\frac{1}{3} x^{2} - 6 x + x - 3))

Simplify

= 4 x^{2} - 6 x

Add a constant to the solution

= 4 x^{2} - 6 x + C

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What is the integral of (4(x-3))/(sqrt(x^2-6x)) ?
The integral of (4(x-3))/(sqrt(x^2-6x)) is 4sqrt(x^2-6x)+C