What is the integral of (2x-9)/(sqrt(x^2-9x+5)) ?

The integral of (2x-9)/(sqrt(x^2-9x+5)) is 2sqrt(x^2-9x+5)+C

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integral of (2x-9)/(sqrt(x^2-9x+5))

\int \frac{2 x - 9}{x ^{2} - 9 x + 5} d x

2 x^{2} - 9 x + 5 + C

Solution steps

\int \frac{2 x - 9}{x ^{2} - 9 x + 5} d x

Expand

\frac{2 x - 9}{x ^{2} - 9 x + 5} : \frac{2 x}{x ^{2} - 9 x + 5} - \frac{9}{x ^{2} - 9 x + 5}

= \int \frac{2 x}{x ^{2} - 9 x + 5} - \frac{9}{x ^{2} - 9 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{2 x}{x ^{2} - 9 x + 5} d x - \int \frac{9}{x ^{2} - 9 x + 5} d x

\int \frac{2 x}{x ^{2} - 9 x + 5} d x = 2 x^{2} - 9 x + 5 + 9 ln \frac{1}{61} (4 x^{2} - 36 x + 20) + \frac{2}{61} x - \frac{9}{61}

\int \frac{9}{x ^{2} - 9 x + 5} d x = 9 ln \frac{1}{61} (4 x^{2} - 36 x + 20) + \frac{2}{61} x - \frac{9}{61}

= 2 x^{2} - 9 x + 5 + 9 ln \frac{1}{61} (4 x^{2} - 36 x + 20) + \frac{2}{61} x - \frac{9}{61} - 9 ln \frac{1}{61} (4 x^{2} - 36 x + 20) + \frac{2}{61} x - \frac{9}{61}

Simplify

= 2 x^{2} - 9 x + 5

Add a constant to the solution

= 2 x^{2} - 9 x + 5 + C

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What is the integral of (2x-9)/(sqrt(x^2-9x+5)) ?
The integral of (2x-9)/(sqrt(x^2-9x+5)) is 2sqrt(x^2-9x+5)+C