What is the integral of (2x-7)/(x^2-12x+32) ?

The integral of (2x-7)/(x^2-12x+32) is -1/4 ln|x-4|+9/4 ln|x-8|+C

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integral of (2x-7)/(x^2-12x+32)

\int \frac{2 x - 7}{x ^{2} - 12 x + 32} d x

- \frac{1}{4} ln ∣ x - 4 ∣ + \frac{9}{4} ln ∣ x - 8 ∣ + C

Solution steps

\int \frac{2 x - 7}{x ^{2} - 12 x + 32} d x

Complete the square

x^{2} - 12 x + 32 : (x - 6)^{2} - 4

= \int \frac{2 x - 7}{( x - 6 ) ^{2} - 4} d x

Apply u-substitution

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

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

Take the partial fraction of

\frac{2 u + 5}{u ^{2} - 4} : - \frac{1}{4 ( u + 2 )} + \frac{9}{4 ( u - 2 )}

= \int - \frac{1}{4 ( u + 2 )} + \frac{9}{4 ( u - 2 )} d u

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{1}{4 ( u + 2 )} d u + \int \frac{9}{4 ( u - 2 )} d u

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

\int \frac{9}{4 ( u - 2 )} d u = \frac{9}{4} ln ∣ u - 2 ∣

= - \frac{1}{4} ln ∣ u + 2 ∣ + \frac{9}{4} ln ∣ u - 2 ∣

Substitute back

u = x - 6

= - \frac{1}{4} ln ∣ x - 6 + 2 ∣ + \frac{9}{4} ln ∣ x - 6 - 2 ∣

Simplify

- \frac{1}{4} ln ∣ x - 6 + 2 ∣ + \frac{9}{4} ln ∣ x - 6 - 2 ∣ : - \frac{1}{4} ln ∣ x - 4 ∣ + \frac{9}{4} ln ∣ x - 8 ∣

= - \frac{1}{4} ln ∣ x - 4 ∣ + \frac{9}{4} ln ∣ x - 8 ∣

Add a constant to the solution

= - \frac{1}{4} ln ∣ x - 4 ∣ + \frac{9}{4} ln ∣ x - 8 ∣ + C

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What is the integral of (2x-7)/(x^2-12x+32) ?
The integral of (2x-7)/(x^2-12x+32) is -1/4 ln|x-4|+9/4 ln|x-8|+C