What is the integral of (x^2-5x+3)/(x^3-4x^2+3x) ?

The integral of (x^2-5x+3)/(x^3-4x^2+3x) is ln|x|+1/2 ln|x-1|-1/2 ln|x-3|+C

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integral of (x^2-5x+3)/(x^3-4x^2+3x)

\int \frac{x ^{2} - 5 x + 3}{x ^{3} - 4 x ^{2} + 3 x} d x

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

Solution steps

\int \frac{x ^{2} - 5 x + 3}{x ^{3} - 4 x ^{2} + 3 x} d x

Take the partial fraction of

\frac{x ^{2} - 5 x + 3}{x ^{3} - 4 x ^{2} + 3 x} : \frac{1}{x} + \frac{1}{2 ( x - 1 )} - \frac{1}{2 ( x - 3 )}

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

Add a constant to the solution

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

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

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

\int (x + 1) cos (x^{2} + 2 x) d x

\int \frac{8 x ^{3}}{4 x ^{2} + 9} d x

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

What is the integral of (x^2-5x+3)/(x^3-4x^2+3x) ?
The integral of (x^2-5x+3)/(x^3-4x^2+3x) is ln|x|+1/2 ln|x-1|-1/2 ln|x-3|+C