What is the integral of (6x^2-8x+5)/(x^3-2x^2+x) ?

The integral of (6x^2-8x+5)/(x^3-2x^2+x) is 5ln|x|+ln|x-1|-3/(x-1)+C

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integral of (6x^2-8x+5)/(x^3-2x^2+x)

\int \frac{6 x ^{2} - 8 x + 5}{x ^{3} - 2 x ^{2} + x} d x

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

Solution steps

\int \frac{6 x ^{2} - 8 x + 5}{x ^{3} - 2 x ^{2} + x} d x

Take the partial fraction of

\frac{6 x ^{2} - 8 x + 5}{x ^{3} - 2 x ^{2} + x} : \frac{5}{x} + \frac{1}{x - 1} + \frac{3}{( x - 1 ) ^{2}}

= \int \frac{5}{x} + \frac{1}{x - 1} + \frac{3}{( 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{5}{x} d x + \int \frac{1}{x - 1} d x + \int \frac{3}{( x - 1 ) ^{2}} d x

\int \frac{5}{x} d x = 5 ln ∣ x ∣

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

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

= 5 ln ∣ x ∣ + ln ∣ x - 1 ∣ - \frac{3}{x - 1}

Add a constant to the solution

= 5 ln ∣ x ∣ + ln ∣ x - 1 ∣ - \frac{3}{x - 1} + C

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What is the integral of (6x^2-8x+5)/(x^3-2x^2+x) ?
The integral of (6x^2-8x+5)/(x^3-2x^2+x) is 5ln|x|+ln|x-1|-3/(x-1)+C