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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

x \to 1 lim (\frac{x ^{4} - 1}{x ^{3} - 1})

3 x - 32 x, at x = 4

y = x^{2} - 12, y = x - 6

\frac{\partial}{\partial x} (x + \frac{x - y}{y - z})

\frac{d}{d x} (tan (cos (\frac{x}{2})))

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