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

The integral of (x^2+2x+3)/(x^2-3x+2) is x-ln|x-1|+4ln|x-2|+ln|4x^2-12x+8|-6ln|2x-2|+6ln|2x-4|+C

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

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

x - ln ∣ x - 1 ∣ + 4 ln ∣ x - 2 ∣ + ln 4 x^{2} - 12 x + 8 - 6 ln ∣ 2 x - 2 ∣ + 6 ln ∣ 2 x - 4 ∣ + C

Solution steps

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

Expand

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

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

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

\int \frac{2 x}{x ^{2} - 3 x + 2} d x = ln 4 x^{2} - 12 x + 8 - 3 ln ∣ 2 x - 2 ∣ + 3 ln ∣ 2 x - 4 ∣

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

= x - ln ∣ x - 1 ∣ + 4 ln ∣ x - 2 ∣ + ln 4 x^{2} - 12 x + 8 - 3 ln ∣ 2 x - 2 ∣ + 3 ln ∣ 2 x - 4 ∣ - 6 (\frac{1}{2} ln ∣ 2 x - 2 ∣ - \frac{1}{2} ln ∣ 2 x - 4 ∣)

Simplify

x - ln ∣ x - 1 ∣ + 4 ln ∣ x - 2 ∣ + ln 4 x^{2} - 12 x + 8 - 3 ln ∣ 2 x - 2 ∣ + 3 ln ∣ 2 x - 4 ∣ - 6 (\frac{1}{2} ln ∣ 2 x - 2 ∣ - \frac{1}{2} ln ∣ 2 x - 4 ∣) : x - ln ∣ x - 1 ∣ + 4 ln ∣ x - 2 ∣ + ln 4 x^{2} - 12 x + 8 - 6 ln ∣ 2 x - 2 ∣ + 6 ln ∣ 2 x - 4 ∣

= x - ln ∣ x - 1 ∣ + 4 ln ∣ x - 2 ∣ + ln 4 x^{2} - 12 x + 8 - 6 ln ∣ 2 x - 2 ∣ + 6 ln ∣ 2 x - 4 ∣

Add a constant to the solution

= x - ln ∣ x - 1 ∣ + 4 ln ∣ x - 2 ∣ + ln 4 x^{2} - 12 x + 8 - 6 ln ∣ 2 x - 2 ∣ + 6 ln ∣ 2 x - 4 ∣ + C

\int \frac{x e ^{2 y}}{4 - y} d x

\int \frac{10}{5 ^{3 x}} d x

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

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

\int \frac{( x ^{5} + 1 )}{( x ^{6} + 6 x )} d x

What is the integral of (x^2+2x+3)/(x^2-3x+2) ?
The integral of (x^2+2x+3)/(x^2-3x+2) is x-ln|x-1|+4ln|x-2|+ln|4x^2-12x+8|-6ln|2x-2|+6ln|2x-4|+C