What is the integral of ((6x-6))/(sqrt(x^2-2x+5)) ?

The integral of ((6x-6))/(sqrt(x^2-2x+5)) is 6sqrt(x^2-2x+5)+C

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integral of ((6x-6))/(sqrt(x^2-2x+5))

\int \frac{( 6 x - 6 )}{x ^{2} - 2 x + 5} d x

6 x^{2} - 2 x + 5 + C

Solution steps

\int \frac{6 x - 6}{x ^{2} - 2 x + 5} d x

Expand

\frac{6 x - 6}{x ^{2} - 2 x + 5} : \frac{6 x}{x ^{2} - 2 x + 5} - \frac{6}{x ^{2} - 2 x + 5}

= \int \frac{6 x}{x ^{2} - 2 x + 5} - \frac{6}{x ^{2} - 2 x + 5} 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{6 x}{x ^{2} - 2 x + 5} d x - \int \frac{6}{x ^{2} - 2 x + 5} d x

\int \frac{6 x}{x ^{2} - 2 x + 5} d x = 6 (x^{2} - 2 x + 5 + ln (\frac{1}{2} x - 1 + x^{2} - 2 x + 5))

\int \frac{6}{x ^{2} - 2 x + 5} d x = 6 ln (\frac{1}{2} x - 1 + x^{2} - 2 x + 5)

= 6 (x^{2} - 2 x + 5 + ln (\frac{1}{2} x - 1 + x^{2} - 2 x + 5)) - 6 ln (\frac{1}{2} x - 1 + x^{2} - 2 x + 5)

Simplify

6 (x^{2} - 2 x + 5 + ln (\frac{1}{2} x - 1 + x^{2} - 2 x + 5)) - 6 ln (\frac{1}{2} x - 1 + x^{2} - 2 x + 5) : 6 x^{2} - 2 x + 5

= 6 x^{2} - 2 x + 5

Add a constant to the solution

= 6 x^{2} - 2 x + 5 + C

What is the integral of ((6x-6))/(sqrt(x^2-2x+5)) ?
The integral of ((6x-6))/(sqrt(x^2-2x+5)) is 6sqrt(x^2-2x+5)+C