What is the integral of (4x+6)/(x^3-5x^2+6x) ?

The integral of (4x+6)/(x^3-5x^2+6x) is (4x)/(x^2-5x+6)-7ln|x-2|+8/(x-2)+6ln|x-3|-(12)/(x-3)+ln|x|+C

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integral of (4x+6)/(x^3-5x^2+6x)

\int \frac{4 x + 6}{x ^{3} - 5 x ^{2} + 6 x} d x

\frac{4 x}{x ^{2} - 5 x + 6} - 7 ln ∣ x - 2 ∣ + \frac{8}{x - 2} + 6 ln ∣ x - 3 ∣ - \frac{12}{x - 3} + ln ∣ x ∣ + C

Solution steps

\int \frac{4 x + 6}{x ^{3} - 5 x ^{2} + 6 x} d x

Expand

\frac{4 x + 6}{x ^{3} - 5 x ^{2} + 6 x} : \frac{4 x}{x ^{3} - 5 x ^{2} + 6 x} + \frac{6}{x ^{3} - 5 x ^{2} + 6 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{4 x}{x ^{3} - 5 x ^{2} + 6 x} d x + \int \frac{6}{x ^{3} - 5 x ^{2} + 6 x} d x

\int \frac{4 x}{x ^{3} - 5 x ^{2} + 6 x} d x = 4 (\frac{x}{x ^{2} - 5 x + 6} - ln ∣ x - 2 ∣ + \frac{2}{x - 2} + ln ∣ x - 3 ∣ - \frac{3}{x - 3})

\int \frac{6}{x ^{3} - 5 x ^{2} + 6 x} d x = ln ∣ x ∣ - 3 ln ∣ x - 2 ∣ + 2 ln ∣ x - 3 ∣

= 4 (\frac{x}{x ^{2} - 5 x + 6} - ln ∣ x - 2 ∣ + \frac{2}{x - 2} + ln ∣ x - 3 ∣ - \frac{3}{x - 3}) + ln ∣ x ∣ - 3 ln ∣ x - 2 ∣ + 2 ln ∣ x - 3 ∣

Simplify

4 (\frac{x}{x ^{2} - 5 x + 6} - ln ∣ x - 2 ∣ + \frac{2}{x - 2} + ln ∣ x - 3 ∣ - \frac{3}{x - 3}) + ln ∣ x ∣ - 3 ln ∣ x - 2 ∣ + 2 ln ∣ x - 3 ∣ : \frac{4 x}{x ^{2} - 5 x + 6} - 7 ln ∣ x - 2 ∣ + \frac{8}{x - 2} + 6 ln ∣ x - 3 ∣ - \frac{12}{x - 3} + ln ∣ x ∣

= \frac{4 x}{x ^{2} - 5 x + 6} - 7 ln ∣ x - 2 ∣ + \frac{8}{x - 2} + 6 ln ∣ x - 3 ∣ - \frac{12}{x - 3} + ln ∣ x ∣

Add a constant to the solution

= \frac{4 x}{x ^{2} - 5 x + 6} - 7 ln ∣ x - 2 ∣ + \frac{8}{x - 2} + 6 ln ∣ x - 3 ∣ - \frac{12}{x - 3} + ln ∣ x ∣ + C

\int x (9 x^{3} - 2 x + \frac{8}{x}) d x

\int x (e^{x} - 1) d x

- x^{2}

\int x e^{0.1 x} d x

\int \frac{10 ln ( x ^{2} )}{x} d x

What is the integral of (4x+6)/(x^3-5x^2+6x) ?
The integral of (4x+6)/(x^3-5x^2+6x) is (4x)/(x^2-5x+6)-7ln|x-2|+8/(x-2)+6ln|x-3|-(12)/(x-3)+ln|x|+C