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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

Add a constant to the solution

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

\frac{d}{d x} (f (x) (g (x)))

\int (4 x + 5)^{3} d x

e^{- \frac{x ^{2}}{2}}

\frac{d}{d x} ((4 x^{3} + 3)^{2})

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