What is the integral of (x+1)/(x^3-6x^2+9x) ?

The integral of (x+1)/(x^3-6x^2+9x) is 1/9 ln|x|-1/9 ln|x-3|-4/(3(x-3))+C

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integral of (x+1)/(x^3-6x^2+9x)

\int \frac{x + 1}{x ^{3} - 6 x ^{2} + 9 x} d x

\frac{1}{9} ln ∣ x ∣ - \frac{1}{9} ln ∣ x - 3 ∣ - \frac{4}{3 ( x - 3 )} + C

Solution steps

\int \frac{x + 1}{x ^{3} - 6 x ^{2} + 9 x} d x

Take the partial fraction of

\frac{x + 1}{x ^{3} - 6 x ^{2} + 9 x} : \frac{1}{9 x} - \frac{1}{9 ( x - 3 )} + \frac{4}{3 ( x - 3 ) ^{2}}

= \int \frac{1}{9 x} - \frac{1}{9 ( x - 3 )} + \frac{4}{3 ( x - 3 ) ^{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}{9 x} d x - \int \frac{1}{9 ( x - 3 )} d x + \int \frac{4}{3 ( x - 3 ) ^{2}} d x

\int \frac{1}{9 x} d x = \frac{1}{9} ln ∣ x ∣

\int \frac{1}{9 ( x - 3 )} d x = \frac{1}{9} ln ∣ x - 3 ∣

\int \frac{4}{3 ( x - 3 ) ^{2}} d x = - \frac{4}{3 ( x - 3 )}

= \frac{1}{9} ln ∣ x ∣ - \frac{1}{9} ln ∣ x - 3 ∣ - \frac{4}{3 ( x - 3 )}

Add a constant to the solution

= \frac{1}{9} ln ∣ x ∣ - \frac{1}{9} ln ∣ x - 3 ∣ - \frac{4}{3 ( x - 3 )} + C

x \to 0 lim (- (\frac{6 x}{x ^{4}}))

\int (4 x + 1) d x

\int x e^{- \frac{x}{10}} d x

\int_{0}^{2} 2 e^{x} d x

(3 t - 1) (2 t - 2)^{- 1}

What is the integral of (x+1)/(x^3-6x^2+9x) ?
The integral of (x+1)/(x^3-6x^2+9x) is 1/9 ln|x|-1/9 ln|x-3|-4/(3(x-3))+C