What is the integral of (2x-9)/(sqrt(x^2-9x+1)) ?

The integral of (2x-9)/(sqrt(x^2-9x+1)) is 2sqrt(x^2-9x+1)+C

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integral of (2x-9)/(sqrt(x^2-9x+1))

\int \frac{2 x - 9}{x ^{2} - 9 x + 1} d x

2 x^{2} - 9 x + 1 + C

Solution steps

\int \frac{2 x - 9}{x ^{2} - 9 x + 1} d x

Expand

\frac{2 x - 9}{x ^{2} - 9 x + 1} : \frac{2 x}{x ^{2} - 9 x + 1} - \frac{9}{x ^{2} - 9 x + 1}

= \int \frac{2 x}{x ^{2} - 9 x + 1} - \frac{9}{x ^{2} - 9 x + 1} 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 + 1} d x - \int \frac{9}{x ^{2} - 9 x + 1} d x

\int \frac{2 x}{x ^{2} - 9 x + 1} d x = 2 x^{2} - 9 x + 1 + 9 ln \frac{1}{77} (4 x^{2} - 36 x + 4) + \frac{2}{77} x - \frac{9}{77}

\int \frac{9}{x ^{2} - 9 x + 1} d x = 9 ln \frac{1}{77} (4 x^{2} - 36 x + 4) + \frac{2}{77} x - \frac{9}{77}

= 2 x^{2} - 9 x + 1 + 9 ln \frac{1}{77} (4 x^{2} - 36 x + 4) + \frac{2}{77} x - \frac{9}{77} - 9 ln \frac{1}{77} (4 x^{2} - 36 x + 4) + \frac{2}{77} x - \frac{9}{77}

Simplify

= 2 x^{2} - 9 x + 1

Add a constant to the solution

= 2 x^{2} - 9 x + 1 + C

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

y = x^{2} + 1, (- 2, 5)

x, at 25

x^{4} - 2 x^{2} + 4

\frac{\partial}{\partial x} (\frac{5 x - 9 y}{5 x + 9 y})

What is the integral of (2x-9)/(sqrt(x^2-9x+1)) ?
The integral of (2x-9)/(sqrt(x^2-9x+1)) is 2sqrt(x^2-9x+1)+C