What is the integral of 1/((x^2-4x-5)) ?

The integral of 1/((x^2-4x-5)) is -1/6 ln|x+1|+1/6 ln|x-5|+C

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integral of 1/((x^2-4x-5))

\int \frac{1}{( x ^{2} - 4 x - 5 )} d x

- \frac{1}{6} ln ∣ x + 1 ∣ + \frac{1}{6} ln ∣ x - 5 ∣ + C

Solution steps

\int \frac{1}{x ^{2} - 4 x - 5} d x

Take the partial fraction of

\frac{1}{x ^{2} - 4 x - 5} : - \frac{1}{6 ( x + 1 )} + \frac{1}{6 ( x - 5 )}

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

\int \frac{1}{6 ( x + 1 )} d x = \frac{1}{6} ln ∣ x + 1 ∣

\int \frac{1}{6 ( x - 5 )} d x = \frac{1}{6} ln ∣ x - 5 ∣

= - \frac{1}{6} ln ∣ x + 1 ∣ + \frac{1}{6} ln ∣ x - 5 ∣

Add a constant to the solution

= - \frac{1}{6} ln ∣ x + 1 ∣ + \frac{1}{6} ln ∣ x - 5 ∣ + C

x^{4} + 2 x, 1

(12 x + x^{2} + 0.05 x^{3})^{^{'}}

\int_{- 1}^{1} e^{y} - (y^{2} - 2) d y

n = 2 \sum \infty \frac{3}{4 ^{n}}

\frac{\partial}{\partial x} (cos^{7} (x^{4} y^{6}))

What is the integral of 1/((x^2-4x-5)) ?
The integral of 1/((x^2-4x-5)) is -1/6 ln|x+1|+1/6 ln|x-5|+C