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

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

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integral of 2/((x^2-9))

integral

\frac{2}{( x ^{2} - 9 )}

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

Solution steps

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

Take the constant out:

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

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

Apply Integral Substitution

= 2 \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

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

Factor

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

= 2 \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

= 2 \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}

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

Substitute back

u = \frac{x}{3}

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

Simplify

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

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

Add a constant to the solution

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

\frac{1}{2 x + 1}

(e^{e})^{x} \cdot e^{2} x

\frac{( 1 - e ^{- x} )}{x}

\frac{sin ( x )}{cos ^{3} ( x )}

\frac{2 x}{( 1 - x ^{2} )}

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