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

The integral of 6/(x^2-9) is -ln| x/3+1|+ln| x/3-1|+C

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integral of 6/(x^2-9)

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

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

Solution steps

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

Take the constant out:

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

= 6 \cdot \int \frac{1}{x ^{2} - 9} d x

Apply Integral Substitution

= 6 \cdot \int \frac{1}{3 ( u ^{2} - 1 )} d u

Take the constant out:

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

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

Factor

u^{2} - 1 : - (- u^{2} + 1)

= 6 \cdot \frac{1}{3} \cdot \int \frac{1}{- ( - u ^{2} + 1 )} d u

Take the constant out:

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

= 6 \cdot \frac{1}{3} (- \int \frac{1}{- u ^{2} + 1} d u)

Use the common integral:

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

= 6 \cdot \frac{1}{3} (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = \frac{x}{3}

= 6 \cdot \frac{1}{3} (- (\frac{ln \frac{x}{3} + 1}{2} - \frac{ln \frac{x}{3} - 1}{2}))

Simplify

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

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

Add a constant to the solution

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

\int x^{\frac{5}{6}} d x

\int \frac{1}{x 8 x + 9} d x

f (x) = \frac{3}{2 x - 1}, (1, 3)

\frac{\partial}{\partial y} (- x y z^{2})

y = x^{2} 25 - x^{2}

What is the integral of 6/(x^2-9) ?
The integral of 6/(x^2-9) is -ln| x/3+1|+ln| x/3-1|+C