What is the integral of 1/(x^3-4x^2+3x) ?

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

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integral of 1/(x^3-4x^2+3x)

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

\frac{1}{3} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{6} ln ∣ x - 3 ∣ + C

Solution steps

\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 )}

= \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

= \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 ∣

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

Add a constant to the solution

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

\int 4 x (x^{2} + 5)^{3} d x

h \to 0 lim (x^{2} + 3 x h)

\int \frac{y ^{2} - 16}{y} d y

f (x) = ln (1 - x^{2})

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

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