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

The integral of (x-1)/(x^2-4x-5) is 1/3 ln|x+1|+2/3 ln|x-5|+C

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

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

\frac{1}{3} ln ∣ x + 1 ∣ + \frac{2}{3} ln ∣ x - 5 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x - 1}{x ^{2} - 4 x - 5} : \frac{1}{3 ( x + 1 )} + \frac{2}{3 ( x - 5 )}

= \int \frac{1}{3 ( x + 1 )} + \frac{2}{3 ( 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}{3 ( x + 1 )} d x + \int \frac{2}{3 ( x - 5 )} d x

\int \frac{1}{3 ( x + 1 )} d x = \frac{1}{3} ln ∣ x + 1 ∣

\int \frac{2}{3 ( x - 5 )} d x = \frac{2}{3} ln ∣ x - 5 ∣

= \frac{1}{3} ln ∣ x + 1 ∣ + \frac{2}{3} ln ∣ x - 5 ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ x + 1 ∣ + \frac{2}{3} ln ∣ x - 5 ∣ + C

\int_{0}^{1} 4 t^{2} + 9 t^{4} d t

\frac{d}{d x} (\frac{1}{x} + x^{2} - 2)

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

- \frac{7}{1 + x ^{2}}

n = 1 \sum \infty \frac{n}{2 n ^{3} + 3}

What is the integral of (x-1)/(x^2-4x-5) ?
The integral of (x-1)/(x^2-4x-5) is 1/3 ln|x+1|+2/3 ln|x-5|+C