What is the integral of 1/((y+1)(y-2)) ?

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

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integral of 1/((y+1)(y-2))

\int \frac{1}{( y + 1 ) ( y - 2 )} d y

- \frac{1}{3} ln ∣ y + 1 ∣ + \frac{1}{3} ln ∣ y - 2 ∣ + C

Solution steps

\int \frac{1}{( y + 1 ) ( y - 2 )} d y

Take the partial fraction of

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

= \int - \frac{1}{3 ( y + 1 )} + \frac{1}{3 ( y - 2 )} 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}{3 ( y + 1 )} d y + \int \frac{1}{3 ( y - 2 )} d y

\int \frac{1}{3 ( y + 1 )} d y = \frac{1}{3} ln ∣ y + 1 ∣

\int \frac{1}{3 ( y - 2 )} d y = \frac{1}{3} ln ∣ y - 2 ∣

= - \frac{1}{3} ln ∣ y + 1 ∣ + \frac{1}{3} ln ∣ y - 2 ∣

Add a constant to the solution

= - \frac{1}{3} ln ∣ y + 1 ∣ + \frac{1}{3} ln ∣ y - 2 ∣ + C

\frac{\partial}{\partial x} (e^{2 x - y})

f (x) = x^{2} + sin (x)

\frac{d}{d u} (arctan (u))

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

x \to \frac{π}{6} lim (tan (x))

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