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

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

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

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

- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 3 ∣ + C

Solution steps

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

Take the partial fraction of

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

= \int - \frac{1}{4 ( x + 1 )} + \frac{1}{4 ( x - 3 )} 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}{4 ( x + 1 )} d x + \int \frac{1}{4 ( x - 3 )} d x

\int \frac{1}{4 ( x + 1 )} d x = \frac{1}{4} ln ∣ x + 1 ∣

\int \frac{1}{4 ( x - 3 )} d x = \frac{1}{4} ln ∣ x - 3 ∣

= - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 3 ∣

Add a constant to the solution

= - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 3 ∣ + C

f (x) = - \frac{1}{4} x^{2}

f^{^{'}} (x) = \frac{x}{4}

\int 121 - t^{2} d t

f (x) = x^{2} - 5, at x = - 4

\int 2 y d y

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