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

The integral of 1/(x^2(x-1)(x-2)) is 3/4 ln|x|-1/(2x)-ln|x-1|+1/4 ln|x-2|+C

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

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

\frac{3}{4} ln ∣ x ∣ - \frac{1}{2 x} - ln ∣ x - 1 ∣ + \frac{1}{4} ln ∣ x - 2 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{x ^{2} ( x - 1 ) ( x - 2 )} : \frac{3}{4 x} + \frac{1}{2 x ^{2}} - \frac{1}{x - 1} + \frac{1}{4 ( x - 2 )}

= \int \frac{3}{4 x} + \frac{1}{2 x ^{2}} - \frac{1}{x - 1} + \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{3}{4 x} d x + \int \frac{1}{2 x ^{2}} d x - \int \frac{1}{x - 1} d x + \int \frac{1}{4 ( x - 2 )} d x

\int \frac{3}{4 x} d x = \frac{3}{4} ln ∣ x ∣

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

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

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

= \frac{3}{4} ln ∣ x ∣ - \frac{1}{2 x} - ln ∣ x - 1 ∣ + \frac{1}{4} ln ∣ x - 2 ∣

Add a constant to the solution

= \frac{3}{4} ln ∣ x ∣ - \frac{1}{2 x} - ln ∣ x - 1 ∣ + \frac{1}{4} ln ∣ x - 2 ∣ + C

\int \frac{x ^{4} + x - 36}{x ^{2} + 6} d x

11 x d y + 11 y d x + x d x = 0

\int_{2}^{4} \frac{1}{x ^{2}} d x

t^{2} y^{^{''}} + t y^{^{'}} + 16 y = 0

\int \frac{( 2 x - 1 ) ( x + 3 )}{6} d x

What is the integral of 1/(x^2(x-1)(x-2)) ?
The integral of 1/(x^2(x-1)(x-2)) is 3/4 ln|x|-1/(2x)-ln|x-1|+1/4 ln|x-2|+C