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))

integral

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

\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

\frac{1}{7 [ \frac{1}{x} ]}

\frac{( 3 t + 4 )}{( 2 t + 3 )}

\frac{1}{( y - 2 )}

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

(4 x + 9)^{\frac{1}{2}}

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