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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (2 z)

\frac{2 x}{x ^{2} + 1}

\int \frac{4 + 6 x - 3 x ^{2}}{x} d x

\int (8 + x^{4})^{3} 4 x^{3} d x

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