What is the integral of 4/((x+2)(x-2)) ?

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

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integral of 4/((x+2)(x-2))

\int \frac{4}{( x + 2 ) ( x - 2 )} d x

- ln ∣ x + 2 ∣ + ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{4}{( x + 2 ) ( x - 2 )} d x

Take the constant out:

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

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

Take the partial fraction of

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

= 4 \cdot \int - \frac{1}{4 ( x + 2 )} + \frac{1}{4 ( x - 2 )} 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

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

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

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

= 4 (- \frac{1}{4} ln ∣ x + 2 ∣ + \frac{1}{4} ln ∣ x - 2 ∣)

Simplify

4 (- \frac{1}{4} ln ∣ x + 2 ∣ + \frac{1}{4} ln ∣ x - 2 ∣) : - ln ∣ x + 2 ∣ + ln ∣ x - 2 ∣

= - ln ∣ x + 2 ∣ + ln ∣ x - 2 ∣

Add a constant to the solution

= - ln ∣ x + 2 ∣ + ln ∣ x - 2 ∣ + C

y^{^{'}} = \frac{2 cos ( 2 x )}{3 + 2 y}, y (0) = - 1

\int_{0}^{1} \frac{x - 1}{x ^{2} + 3 x + 2} d x

\frac{\partial}{\partial x} (ln (x + 7 y + 8 z))

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

f (x) = \frac{e ^{x}}{x}, at x = 1

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