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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \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 )}

= 3 \cdot \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

= 3 (- \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 ∣

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

Add a constant to the solution

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

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

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

\int_{0}^{x} 1 + t^{2} d t

\frac{d}{d x} (4 x^{\frac{1}{4}})

x \to 3 lim (x - 1)

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