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

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

\frac{1}{9 s + 1}

\frac{d}{d x} (3 x^{4} - 5 x^{3} + 2 x^{2})

y^{^{''}} - 5 y^{^{'}} + 6 y = e^{2 t}

x \to 3 lim (\frac{1}{( x - 3 ) ^{2}})

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