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 ∣

Simplify

= - \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

\int_{0}^{3} 3 x^{3} d x

\frac{1}{2 + t}

\int 8 csc (x) d x

f (x) = 5 x + 39, at x = 5

\frac{x}{5 + x ^{2}}

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