What is the integral of (3x)/(x^2-6x+9) ?

The integral of (3x)/(x^2-6x+9) is 3(ln|x-3|-3/(x-3))+C

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integral of (3x)/(x^2-6x+9)

\int \frac{3 x}{x ^{2} - 6 x + 9} d x

3 (ln ∣ x - 3 ∣ - \frac{3}{x - 3}) + C

Solution steps

\int \frac{3 x}{x ^{2} - 6 x + 9} d x

Take the constant out:

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

= 3 \cdot \int \frac{x}{x ^{2} - 6 x + 9} d x

Complete the square

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

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

Apply u-substitution

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

= 3 \cdot \int \frac{u + 3}{u ^{2}} d u

Expand

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

= 3 \cdot \int \frac{1}{u} + \frac{3}{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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

= 3 (ln ∣ u ∣ - \frac{3}{u})

Substitute back

u = x - 3

= 3 (ln ∣ x - 3 ∣ - \frac{3}{x - 3})

Add a constant to the solution

= 3 (ln ∣ x - 3 ∣ - \frac{3}{x - 3}) + C

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What is the integral of (3x)/(x^2-6x+9) ?
The integral of (3x)/(x^2-6x+9) is 3(ln|x-3|-3/(x-3))+C