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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

Add a constant to the solution

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

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