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

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

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

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

x - ln ∣ x - 1 ∣ + 4 ln ∣ x - 2 ∣ - 4 (\frac{1}{2} ln ∣ 2 x - 2 ∣ - \frac{1}{2} ln ∣ 2 x - 4 ∣) + C

Solution steps

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

Expand

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

= x - ln ∣ x - 1 ∣ + 4 ln ∣ x - 2 ∣ - 4 (\frac{1}{2} ln ∣ 2 x - 2 ∣ - \frac{1}{2} ln ∣ 2 x - 4 ∣)

Add a constant to the solution

= x - ln ∣ x - 1 ∣ + 4 ln ∣ x - 2 ∣ - 4 (\frac{1}{2} ln ∣ 2 x - 2 ∣ - \frac{1}{2} ln ∣ 2 x - 4 ∣) + C

\int e^{\frac{- x}{5}} d x

\int 5 (2 - a w^{2})^{3} w d w

\int x (2 x + 1)^{5} d x

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

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

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