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

The integral of 2/(x(x-2)^2) is 1/2 ln|x|-1/2 ln|x-2|-1/(x-2)+C

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

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

\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣ - \frac{1}{x - 2} + C

Solution steps

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

Take the constant out:

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

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

Take the partial fraction of

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

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

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

\int \frac{1}{4 x} d x = \frac{1}{4} ln ∣ x ∣

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

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

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

Simplify

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

= \frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣ - \frac{1}{x - 2}

Add a constant to the solution

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

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\int sin^{3} (4 x) cos^{- 2} (4 x) d x

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