What is the integral of 1/((9-x^2)) ?

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

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integral of 1/((9-x^2))

\int \frac{1}{( 9 - x ^{2} )} d x

\frac{1}{6} (ln \frac{x}{3} + 1 - ln \frac{x}{3} - 1) + C

Solution steps

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

Apply Integral Substitution

= \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

= \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}

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

Substitute back

u = \frac{x}{3}

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

Simplify

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

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

Add a constant to the solution

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

\int - 4 x sin (5 x) d x

y = 3 x + 13

x \to \infty lim (9 x^{2})

6 lo g_{9} (x)

5 cos (9 x), 5 - 5 cos (9 x), [0, \frac{π}{9}]

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