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

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

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

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

\frac{3}{2} ln 4 x^{2} - 12 x + 8 - \frac{1}{2} ln ∣ 2 x - 2 ∣ + \frac{1}{2} ln ∣ 2 x - 4 ∣ + C

Solution steps

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

Expand

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

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

\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 ∣)

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

Simplify

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

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

Add a constant to the solution

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

x + y^{2} = 42, x + y = 0

x = 25 - y^{2}, x = y^{2} - 25

\frac{d}{d x} (1 0^{- 6} x^{3})

f^{^{'}} (3) - \frac{3}{8 ( x + 4 ) ^{4}}

\frac{d}{d x} (x ln (x + 1))

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