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

The integral of ((x^2+x-6)/(x^2-4x+3)) is x-1/2 ln|x-1|+1/2 ln|x-3|+5/2 ln|x^2-4x+3|+C

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

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

x - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ x - 3 ∣ + \frac{5}{2} ln x^{2} - 4 x + 3 + C

Solution steps

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

Expand

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

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

\int \frac{x ^{2}}{x ^{2} - 4 x + 3} d x = 2 ln x^{2} - 4 x + 3 - 5 (\frac{1}{2} ln ∣ x - 1 ∣ - \frac{1}{2} ln ∣ x - 3 ∣) + x

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

\int \frac{6}{x ^{2} - 4 x + 3} d x = 6 (- \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ x - 3 ∣)

= 2 ln x^{2} - 4 x + 3 - 5 (\frac{1}{2} ln ∣ x - 1 ∣ - \frac{1}{2} ln ∣ x - 3 ∣) + x + \frac{1}{2} ln x^{2} - 4 x + 3 - ln ∣ x - 1 ∣ + ln ∣ x - 3 ∣ - 6 (- \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ x - 3 ∣)

Simplify

2 ln x^{2} - 4 x + 3 - 5 (\frac{1}{2} ln ∣ x - 1 ∣ - \frac{1}{2} ln ∣ x - 3 ∣) + x + \frac{1}{2} ln x^{2} - 4 x + 3 - ln ∣ x - 1 ∣ + ln ∣ x - 3 ∣ - 6 (- \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ x - 3 ∣) : x - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ x - 3 ∣ + \frac{5}{2} ln x^{2} - 4 x + 3

= x - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ x - 3 ∣ + \frac{5}{2} ln x^{2} - 4 x + 3

Add a constant to the solution

= x - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ x - 3 ∣ + \frac{5}{2} ln x^{2} - 4 x + 3 + C

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\int \frac{x - 6}{x} d x

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

What is the integral of ((x^2+x-6)/(x^2-4x+3)) ?
The integral of ((x^2+x-6)/(x^2-4x+3)) is x-1/2 ln|x-1|+1/2 ln|x-3|+5/2 ln|x^2-4x+3|+C