What is the integral of (2x-5)/(x^2-5x-1) ?

The integral of (2x-5)/(x^2-5x-1) is ln|4x^2-20x-4|+C

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integral of (2x-5)/(x^2-5x-1)

\int \frac{2 x - 5}{x ^{2} - 5 x - 1} d x

ln 4 x^{2} - 20 x - 4 + C

Solution steps

\int \frac{2 x - 5}{x ^{2} - 5 x - 1} d x

Expand

\frac{2 x - 5}{x ^{2} - 5 x - 1} : \frac{2 x}{x ^{2} - 5 x - 1} - \frac{5}{x ^{2} - 5 x - 1}

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} - 5 x - 1} d x - \int \frac{5}{x ^{2} - 5 x - 1} d x

\int \frac{2 x}{x ^{2} - 5 x - 1} d x = ln 4 x^{2} - 20 x - 4 - \frac{5}{29} (ln \frac{2 x - 5}{29} + 1 - ln \frac{2 x - 5}{29} - 1)

\int \frac{5}{x ^{2} - 5 x - 1} d x = - \frac{5}{29} (ln \frac{2 x - 5}{29} + 1 - ln \frac{2 x - 5}{29} - 1)

= ln 4 x^{2} - 20 x - 4 - \frac{5}{29} (ln \frac{2 x - 5}{29} + 1 - ln \frac{2 x - 5}{29} - 1) - (- \frac{5}{29} (ln \frac{2 x - 5}{29} + 1 - ln \frac{2 x - 5}{29} - 1))

Simplify

ln 4 x^{2} - 20 x - 4 - \frac{5}{29} (ln \frac{2 x - 5}{29} + 1 - ln \frac{2 x - 5}{29} - 1) - (- \frac{5}{29} (ln \frac{2 x - 5}{29} + 1 - ln \frac{2 x - 5}{29} - 1)) : ln 4 x^{2} - 20 x - 4

= ln 4 x^{2} - 20 x - 4

Add a constant to the solution

= ln 4 x^{2} - 20 x - 4 + C

\frac{\partial}{\partial x} (sin (x y) + x cos (y))

\frac{d}{d x} (\frac{x - x}{x ^{\frac{1}{9}}})

y = 2 x + 5, y = 14 - x^{2}, (- 1, 2)

y = x^{\frac{1}{2}}, y = x^{2}

\int_{1}^{e} \frac{1 - ln ( x )}{x} d x

What is the integral of (2x-5)/(x^2-5x-1) ?
The integral of (2x-5)/(x^2-5x-1) is ln|4x^2-20x-4|+C