What is the integral of (16)/((x+2)(x^2-4)) ?

The integral of (16)/((x+2)(x^2-4)) is -ln|x+2|+4/(x+2)+ln|x-2|+C

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integral of (16)/((x+2)(x^2-4))

\int \frac{16}{( x + 2 ) ( x ^{2} - 4 )} d x

- ln ∣ x + 2 ∣ + \frac{4}{x + 2} + ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{16}{( x + 2 ) ( x ^{2} - 4 )} d x

Take the constant out:

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

= 16 \cdot \int \frac{1}{( x + 2 ) ( x ^{2} - 4 )} d x

Take the partial fraction of

\frac{1}{( x + 2 ) ( x ^{2} - 4 )} : - \frac{1}{16 ( x + 2 )} - \frac{1}{4 ( x + 2 ) ^{2}} + \frac{1}{16 ( x - 2 )}

= 16 \cdot \int - \frac{1}{16 ( x + 2 )} - \frac{1}{4 ( x + 2 ) ^{2}} + \frac{1}{16 ( 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

= 16 (- \int \frac{1}{16 ( x + 2 )} d x - \int \frac{1}{4 ( x + 2 ) ^{2}} d x + \int \frac{1}{16 ( x - 2 )} d x)

\int \frac{1}{16 ( x + 2 )} d x = \frac{1}{16} ln ∣ x + 2 ∣

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

\int \frac{1}{16 ( x - 2 )} d x = \frac{1}{16} ln ∣ x - 2 ∣

= 16 (- \frac{1}{16} ln ∣ x + 2 ∣ - (- \frac{1}{4 ( x + 2 )}) + \frac{1}{16} ln ∣ x - 2 ∣)

Simplify

16 (- \frac{1}{16} ln ∣ x + 2 ∣ - (- \frac{1}{4 ( x + 2 )}) + \frac{1}{16} ln ∣ x - 2 ∣) : - ln ∣ x + 2 ∣ + \frac{4}{x + 2} + ln ∣ x - 2 ∣

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

Add a constant to the solution

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

\int e^{6 x} sin (6 x) d x

\frac{d y}{d x} - y^{2} = 0

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

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

What is the integral of (16)/((x+2)(x^2-4)) ?
The integral of (16)/((x+2)(x^2-4)) is -ln|x+2|+4/(x+2)+ln|x-2|+C