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

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

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

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

\frac{1}{4} ln ∣ y - 2 ∣ - \frac{1}{4} ln ∣ y + 2 ∣ + C

Solution steps

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

Take the partial fraction of

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

= \int \frac{1}{4 ( y - 2 )} - \frac{1}{4 ( 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}{4 ( y - 2 )} d y - \int \frac{1}{4 ( y + 2 )} d y

\int \frac{1}{4 ( y - 2 )} d y = \frac{1}{4} ln ∣ y - 2 ∣

\int \frac{1}{4 ( y + 2 )} d y = \frac{1}{4} ln ∣ y + 2 ∣

= \frac{1}{4} ln ∣ y - 2 ∣ - \frac{1}{4} ln ∣ y + 2 ∣

Add a constant to the solution

= \frac{1}{4} ln ∣ y - 2 ∣ - \frac{1}{4} ln ∣ y + 2 ∣ + C

\frac{d}{d x} (x - x)

\int e^{x} (3 e^{x} + 1)^{- \frac{3}{2}} d x

\int z z - 6 d z

y^{^{''}} + (\frac{2}{x}) = 0

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

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