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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

= ln ∣ x ∣ - ln ∣ x - 1 ∣ - \frac{1}{x - 1} - \frac{3}{x - 1} - (ln ∣ x - 1 ∣ - \frac{1}{x - 1})

Simplify

ln ∣ x ∣ - ln ∣ x - 1 ∣ - \frac{1}{x - 1} - \frac{3}{x - 1} - (ln ∣ x - 1 ∣ - \frac{1}{x - 1}) : ln ∣ x ∣ - 2 ln ∣ x - 1 ∣ - \frac{3}{x - 1}

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

Add a constant to the solution

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

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