What is the integral of 6/(x^3-4x^2+3x) ?

The integral of 6/(x^3-4x^2+3x) is 2ln|x|-3ln|x-1|+ln|x-3|+C

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integral of 6/(x^3-4x^2+3x)

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

2 ln ∣ x ∣ - 3 ln ∣ x - 1 ∣ + ln ∣ x - 3 ∣ + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int \frac{1}{x ^{3} - 4 x ^{2} + 3 x} d x

Take the partial fraction of

\frac{1}{x ^{3} - 4 x ^{2} + 3 x} : \frac{1}{3 x} - \frac{1}{2 ( x - 1 )} + \frac{1}{6 ( x - 3 )}

= 6 \cdot \int \frac{1}{3 x} - \frac{1}{2 ( x - 1 )} + \frac{1}{6 ( 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

= 6 (\int \frac{1}{3 x} d x - \int \frac{1}{2 ( x - 1 )} d x + \int \frac{1}{6 ( x - 3 )} d x)

\int \frac{1}{3 x} d x = \frac{1}{3} ln ∣ x ∣

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

\int \frac{1}{6 ( x - 3 )} d x = \frac{1}{6} ln ∣ x - 3 ∣

= 6 (\frac{1}{3} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{6} ln ∣ x - 3 ∣)

Simplify

6 (\frac{1}{3} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{6} ln ∣ x - 3 ∣) : 2 ln ∣ x ∣ - 3 ln ∣ x - 1 ∣ + ln ∣ x - 3 ∣

= 2 ln ∣ x ∣ - 3 ln ∣ x - 1 ∣ + ln ∣ x - 3 ∣

Add a constant to the solution

= 2 ln ∣ x ∣ - 3 ln ∣ x - 1 ∣ + ln ∣ x - 3 ∣ + C

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

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

\int e \cdot e^{e x} d x

\int \frac{1}{( 1 + x ^{2} ) ^{4}} d x

What is the integral of 6/(x^3-4x^2+3x) ?
The integral of 6/(x^3-4x^2+3x) is 2ln|x|-3ln|x-1|+ln|x-3|+C