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

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

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

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

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

Solution steps

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

Expand

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Simplify

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

Add a constant to the solution

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

y = x (x - 14)

\int_{2 y^{2} + 2 z^{2}}^{2} x d x

\frac{1}{1 - x ^{2}} \frac{d y}{d x} + \frac{1}{1 - x} y = e^{- \frac{1}{2} x^{2}}

\int x (2 + x x)^{5} d x

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

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