What is the integral of 1/(x^2(x+2)) ?

The integral of 1/(x^2(x+2)) is -1/4 ln|x|-1/(2x)+1/4 ln|x+2|+C

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integral of 1/(x^2(x+2))

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

- \frac{1}{4} ln ∣ x ∣ - \frac{1}{2 x} + \frac{1}{4} ln ∣ x + 2 ∣ + C

Solution steps

\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 )}

= \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

= - \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 ∣

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

Add a constant to the solution

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

3 x^{2} + 8 x + 2

x \to 0 lim (\frac{2 ( x + h ( x ) ) - 2 x}{h ( x )})

y^{^{''}} + y = 0, y (0) = 1, y^{^{'}} (0) = 1

5 t^{3}

What is the integral of 1/(x^2(x+2)) ?
The integral of 1/(x^2(x+2)) is -1/4 ln|x|-1/(2x)+1/4 ln|x+2|+C