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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

\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 )}

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

Simplify

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

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

Add a constant to the solution

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

\frac{( ( \frac{d y}{d x} ) - e ^{- x} + 5 )}{y} = - 5

\frac{d}{d x} (x^{0.3})

\int 2 cos^{2} (3 x) d x

x \to \infty lim (\frac{9}{2})

y = 8^{4^{x^{2}}}

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