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

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

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

integral

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

x - ln ∣ x - 1 ∣ + 4 ln ∣ x - 2 ∣ - 8 (\frac{1}{2} ln ∣ 2 x - 2 ∣ - \frac{1}{2} ln ∣ 2 x - 4 ∣) + C

Solution steps

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

Expand

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

= x - ln ∣ x - 1 ∣ + 4 ln ∣ x - 2 ∣ - 8 (\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 ∣ - 8 (\frac{1}{2} ln ∣ 2 x - 2 ∣ - \frac{1}{2} ln ∣ 2 x - 4 ∣) + C

\frac{e}{e ^{(x)^{2}}}

\frac{( x - 2 )}{( 2 x + 2 )}

\frac{x ^{3}}{e ^{x}}

sin^{2} (x) cos^{4} (x)

\frac{( x ^{2} + 1 )}{( 3 x - x ^{2} )}

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