What is the integral of (3x-2)/(x^2-5x+6) ?

The integral of (3x-2)/(x^2-5x+6) is

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integral of (3x-2)/(x^2-5x+6)

\int \frac{3 x - 2}{x ^{2} - 5 x + 6} d x

4 (\frac{3}{8} ln 4 x^{2} - 20 x + 24 - \frac{15}{8} (ln ∣ 2 x - 4 ∣ - ln ∣ 2 x - 6 ∣) + \frac{1}{2} ln ∣ 2 x - 4 ∣ - \frac{1}{2} ln ∣ 2 x - 6 ∣) + C

Solution steps

\int \frac{3 x - 2}{x ^{2} - 5 x + 6} d x

Complete the square

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

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

Apply u-substitution

: \int \frac{4 ( 3 ( u + \frac{5}{2} ) - 2 )}{4 u ^{2} - 1} d u

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int \frac{3 ( u + \frac{5}{2} ) - 2}{4 u ^{2} - 1} d u

Expand

\frac{3 ( u + \frac{5}{2} ) - 2}{4 u ^{2} - 1} : \frac{3 ( u + \frac{5}{2} )}{4 u ^{2} - 1} - \frac{2}{4 u ^{2} - 1}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{3 ( u + \frac{5}{2} )}{4 u ^{2} - 1} d u = 3 (\frac{1}{8} ln 4 u^{2} - 1 - \frac{5}{8} (ln ∣ 2 u + 1 ∣ - ln ∣ 2 u - 1 ∣))

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

= 4 (3 (\frac{1}{8} ln 4 u^{2} - 1 - \frac{5}{8} (ln ∣ 2 u + 1 ∣ - ln ∣ 2 u - 1 ∣)) - (- \frac{1}{2} ln ∣ 2 u + 1 ∣ + \frac{1}{2} ln ∣ 2 u - 1 ∣))

Substitute back

u = x - \frac{5}{2}

= 4 (3 (\frac{1}{8} ln 4 (x - \frac{5}{2})^{2} - 1 - \frac{5}{8} (ln 2 (x - \frac{5}{2}) + 1 - ln 2 (x - \frac{5}{2}) - 1)) - (- \frac{1}{2} ln 2 (x - \frac{5}{2}) + 1 + \frac{1}{2} ln 2 (x - \frac{5}{2}) - 1))

Simplify

4 (3 (\frac{1}{8} ln 4 (x - \frac{5}{2})^{2} - 1 - \frac{5}{8} (ln 2 (x - \frac{5}{2}) + 1 - ln 2 (x - \frac{5}{2}) - 1)) - (- \frac{1}{2} ln 2 (x - \frac{5}{2}) + 1 + \frac{1}{2} ln 2 (x - \frac{5}{2}) - 1)) : 4 (\frac{3}{8} ln 4 x^{2} - 20 x + 24 - \frac{15}{8} (ln ∣ 2 x - 4 ∣ - ln ∣ 2 x - 6 ∣) + \frac{1}{2} ln ∣ 2 x - 4 ∣ - \frac{1}{2} ln ∣ 2 x - 6 ∣)

= 4 (\frac{3}{8} ln 4 x^{2} - 20 x + 24 - \frac{15}{8} (ln ∣ 2 x - 4 ∣ - ln ∣ 2 x - 6 ∣) + \frac{1}{2} ln ∣ 2 x - 4 ∣ - \frac{1}{2} ln ∣ 2 x - 6 ∣)

Add a constant to the solution

= 4 (\frac{3}{8} ln 4 x^{2} - 20 x + 24 - \frac{15}{8} (ln ∣ 2 x - 4 ∣ - ln ∣ 2 x - 6 ∣) + \frac{1}{2} ln ∣ 2 x - 4 ∣ - \frac{1}{2} ln ∣ 2 x - 6 ∣) + C

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What is the integral of (3x-2)/(x^2-5x+6) ?
The integral of (3x-2)/(x^2-5x+6) is