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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

Add a constant to the solution

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

\int_{2}^{x} sin (t^{4}) d t

\int_{0}^{3} ln (x^{2} + 1) d x

t \to 1 lim (\frac{sin ( 3 t ) - 3}{t})

x \to \infty lim ((9)^{x})

f (t) = 4 t^{2} cos (2 t) + 5 t e^{- 7 t}

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