What is the integral of 1/(x^3(x-1)^2) ?

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

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integral of 1/(x^3(x-1)^2)

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{3}{x} d x = 3 ln ∣ x ∣

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

\int \frac{1}{x ^{3}} d x = - \frac{1}{2 x ^{2}}

\int \frac{3}{x - 1} d x = 3 ln ∣ x - 1 ∣

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

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

Add a constant to the solution

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

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