What is the integral of (x-1)/(x^2-7x+12) ?

The integral of (x-1)/(x^2-7x+12) is 2(1/4 ln|4x^2-28x+48|-5/4 (ln|2x-6|-ln|2x-8|))+C

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integral of (x-1)/(x^2-7x+12)

\int \frac{x - 1}{x ^{2} - 7 x + 12} d x

2 (\frac{1}{4} ln 4 x^{2} - 28 x + 48 - \frac{5}{4} (ln ∣ 2 x - 6 ∣ - ln ∣ 2 x - 8 ∣)) + C

Solution steps

\int \frac{x - 1}{x ^{2} - 7 x + 12} d x

Complete the square

x^{2} - 7 x + 12 : (x - \frac{7}{2})^{2} - \frac{1}{4}

= \int \frac{x - 1}{( x - \frac{7}{2} ) ^{2} - \frac{1}{4}} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

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

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

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

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

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

Substitute back

u = x - \frac{7}{2}

= 2 (\frac{1}{4} ln 4 (x - \frac{7}{2})^{2} - 1 - \frac{5}{4} (ln 2 (x - \frac{7}{2}) + 1 - ln 2 (x - \frac{7}{2}) - 1))

Simplify

2 (\frac{1}{4} ln 4 (x - \frac{7}{2})^{2} - 1 - \frac{5}{4} (ln 2 (x - \frac{7}{2}) + 1 - ln 2 (x - \frac{7}{2}) - 1)) : 2 (\frac{1}{4} ln 4 x^{2} - 28 x + 48 - \frac{5}{4} (ln ∣ 2 x - 6 ∣ - ln ∣ 2 x - 8 ∣))

= 2 (\frac{1}{4} ln 4 x^{2} - 28 x + 48 - \frac{5}{4} (ln ∣ 2 x - 6 ∣ - ln ∣ 2 x - 8 ∣))

Add a constant to the solution

= 2 (\frac{1}{4} ln 4 x^{2} - 28 x + 48 - \frac{5}{4} (ln ∣ 2 x - 6 ∣ - ln ∣ 2 x - 8 ∣)) + C

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What is the integral of (x-1)/(x^2-7x+12) ?
The integral of (x-1)/(x^2-7x+12) is 2(1/4 ln|4x^2-28x+48|-5/4 (ln|2x-6|-ln|2x-8|))+C