What is the integral of (5x-1)/(x^2-3x-40) ?

The integral of (5x-1)/(x^2-3x-40) is 2ln|x+5|+3ln|x-8|+C

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integral of (5x-1)/(x^2-3x-40)

\int \frac{5 x - 1}{x ^{2} - 3 x - 40} d x

2 ln ∣ x + 5 ∣ + 3 ln ∣ x - 8 ∣ + C

Solution steps

\int \frac{5 x - 1}{x ^{2} - 3 x - 40} d x

Expand

\frac{5 x - 1}{x ^{2} - 3 x - 40} : \frac{5 x}{x ^{2} - 3 x - 40} - \frac{1}{x ^{2} - 3 x - 40}

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{5 x}{x ^{2} - 3 x - 40} d x - \int \frac{1}{x ^{2} - 3 x - 40} d x

\int \frac{5 x}{x ^{2} - 3 x - 40} d x = 5 (\frac{5}{13} ln ∣ x + 5 ∣ + \frac{8}{13} ln ∣ x - 8 ∣)

\int \frac{1}{x ^{2} - 3 x - 40} d x = - \frac{1}{13} ln ∣ x + 5 ∣ + \frac{1}{13} ln ∣ x - 8 ∣

= 5 (\frac{5}{13} ln ∣ x + 5 ∣ + \frac{8}{13} ln ∣ x - 8 ∣) - (- \frac{1}{13} ln ∣ x + 5 ∣ + \frac{1}{13} ln ∣ x - 8 ∣)

Simplify

5 (\frac{5}{13} ln ∣ x + 5 ∣ + \frac{8}{13} ln ∣ x - 8 ∣) - (- \frac{1}{13} ln ∣ x + 5 ∣ + \frac{1}{13} ln ∣ x - 8 ∣) : 2 ln ∣ x + 5 ∣ + 3 ln ∣ x - 8 ∣

= 2 ln ∣ x + 5 ∣ + 3 ln ∣ x - 8 ∣

Add a constant to the solution

= 2 ln ∣ x + 5 ∣ + 3 ln ∣ x - 8 ∣ + C

f (x) = \frac{x}{8 - x}, at x = 3

\frac{\partial}{\partial x} (2 x e^{\frac{x}{2} + y} + 2)

\int \frac{1}{( 6 + x ) ^{3}} d x

f (x) = \frac{1}{1 + x ^{2}}, (- 1, \frac{1}{2})

What is the integral of (5x-1)/(x^2-3x-40) ?
The integral of (5x-1)/(x^2-3x-40) is 2ln|x+5|+3ln|x-8|+C