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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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