What is the integral of (x-3)/(x^2-12x+36) ?

The integral of (x-3)/(x^2-12x+36) is ln|x-6|-3/(x-6)+C

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integral of (x-3)/(x^2-12x+36)

\int \frac{x - 3}{x ^{2} - 12 x + 36} d x

ln ∣ x - 6 ∣ - \frac{3}{x - 6} + C

Solution steps

\int \frac{x - 3}{x ^{2} - 12 x + 36} d x

Apply u-substitution

: \int \frac{u}{u ^{2} - 6 u + 9} d u

= \int \frac{u}{u ^{2} - 6 u + 9} d u

Complete the square

u^{2} - 6 u + 9 : (u - 3)^{2}

= \int \frac{u}{( u - 3 ) ^{2}} d u

Apply u-substitution

: \int \frac{v + 3}{v ^{2}} d v

= \int \frac{v + 3}{v ^{2}} d v

Expand

\frac{v + 3}{v ^{2}} : \frac{1}{v} + \frac{3}{v ^{2}}

= \int \frac{1}{v} + \frac{3}{v ^{2}} d v

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}{v} d v + \int \frac{3}{v ^{2}} d v

\int \frac{1}{v} d v = ln ∣ v ∣

\int \frac{3}{v ^{2}} d v = - \frac{3}{v}

= ln ∣ v ∣ - \frac{3}{v}

Substitute back

= ln ∣ x - 3 - 3 ∣ - \frac{3}{x - 3 - 3}

Simplify

= ln ∣ x - 6 ∣ - \frac{3}{x - 6}

Add a constant to the solution

= ln ∣ x - 6 ∣ - \frac{3}{x - 6} + C

n = 1 \sum \infty \frac{n}{n + 3}

f (x) = (2 + 4 x) (5 - 4 x), at x = 1

y = 2 x^{3} - 2 x^{2} - 2 x

\frac{d y}{d x} + \frac{y}{x} = x^{3}

\frac{d}{d x} (\frac{e ^{a x^{2} + b x + c}}{x})

What is the integral of (x-3)/(x^2-12x+36) ?
The integral of (x-3)/(x^2-12x+36) is ln|x-6|-3/(x-6)+C