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

The integral of 4/(x^2(x-2)) is -ln|x|+2/x+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 ∣ + \frac{2}{x} + 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} - \frac{1}{2 x ^{2}} + \frac{1}{4 ( x - 2 )}

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\int (x^{4} + x + 9) d x

x \to - 1 lim (3 x - 5)

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

\frac{s ^{2} + 1}{s ^{2} ( s + 2 )}

\int 2 x (3 x - 10)^{4} d x

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