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

The integral of 3/(6x-x^2) is 1/2 ln|(x-3)/3+1|-1/2 ln|(x-3)/3-1|+C

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

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

\frac{1}{2} ln \frac{x - 3}{3} + 1 - \frac{1}{2} ln \frac{x - 3}{3} - 1 + C

Solution steps

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

Take the constant out:

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

= 3 \cdot \int \frac{1}{6 x - x ^{2}} d x

Complete the square

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

= 3 \cdot \int \frac{1}{- ( x - 3 ) ^{2} + 9} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 9} d u

= 3 \cdot \int \frac{1}{- u ^{2} + 9} d u

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

= 3 \cdot \frac{1}{3} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= 3 \cdot \frac{1}{3} (\frac{ln \frac{x - 3}{3} + 1}{2} - \frac{ln \frac{x - 3}{3} - 1}{2})

Simplify

3 \cdot \frac{1}{3} (\frac{ln \frac{x - 3}{3} + 1}{2} - \frac{ln \frac{x - 3}{3} - 1}{2}) : \frac{1}{2} ln \frac{x - 3}{3} + 1 - \frac{1}{2} ln \frac{x - 3}{3} - 1

= \frac{1}{2} ln \frac{x - 3}{3} + 1 - \frac{1}{2} ln \frac{x - 3}{3} - 1

Add a constant to the solution

= \frac{1}{2} ln \frac{x - 3}{3} + 1 - \frac{1}{2} ln \frac{x - 3}{3} - 1 + C

\frac{d}{d x} (x^{3} cos (x) - 9 x sin (x) - 9 cos (x))

x \to 0 lim (\frac{0}{x ^{4}})

- 3 x^{\frac{3}{2}} - 1

f (x) = \frac{9}{x}

\frac{d}{d x} (\frac{1}{x} + x - \frac{1}{x} - x)

What is the integral of 3/(6x-x^2) ?
The integral of 3/(6x-x^2) is 1/2 ln|(x-3)/3+1|-1/2 ln|(x-3)/3-1|+C