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

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

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integral of 6/(9-x^2)

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

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

Solution steps

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

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

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t (x) v^{''} + v^{'} = 0

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