The image shows the following question:

What is $f(x)$ if $f'(x) = \frac{9}{9 - x^2}?$

To find $f(x)$, we need to integrate $f'(x)$. Let's solve this:

$f(x) = \int \frac{9}{9 - x^2} \, dx$

Notice that the denominator is a difference of squares, so we can use partial fraction decomposition. We rewrite:

$\frac{9}{9 - x^2} = \frac{9}{(3 - x)(3 + x)}$

The partial fraction decomposition for this is:

$\frac{9}{(3 - x)(3 + x)} = \frac{A}{3 - x} + \frac{B}{3 + x}$

Multiplying both sides by $(3 - x)(3 + x)$ gives:

$9 = A(3 + x) + B(3 - x)$

Expanding both terms:

$9 = A(3 + x) + B(3 - x) = A3 + Ax + B3 - Bx = (A + B)3 + (A - B)x$

Equating coefficients:

$A + B = 0 \quad \text{and} \quad 3A = 9$

From $3A = 9$, we get $A = 3$. Using $A + B = 0$, we find $B = -3$.

Thus, the partial fractions are:

$\frac{9}{(3 - x)(3 + x)} = \frac{3}{3 - x} - \frac{3}{3 + x}$

Now, we integrate each term separately:

$f(x) = \int \left( \frac{3}{3 - x} - \frac{3}{3 + x} \right) dx$

The integrals are straightforward logarithmic forms:

$f(x) = 3 \ln |3 - x| - 3 \ln |3 + x| + C$

We can combine the logarithms:

$f(x) = 3 \ln \left| \frac{3 - x}{3 + x} \right| + C$

Thus, the solution is:

$f(x) = 3 \ln \left| \frac{3 - x}{3 + x} \right| + C$

Would you like any details on the steps, or have any questions about this solution?

Here are some related questions for further exploration:

1. What would the solution look like if $f'(x) = \frac{1}{9 - x^2}$?
2. How do we compute the integral of $f'(x)$ when it involves different forms like $\frac{1}{x^2 + a^2}$?
3. How can partial fraction decomposition help in integrating rational functions?
4. What is the geometric interpretation of the function $f(x) = 3 \ln \left| \frac{3 - x}{3 + x} \right| + C$?
5. How does the constant of integration $C$ affect the shape of the solution?

Tip: When integrating rational functions with quadratic factors, partial fraction decomposition is a powerful tool to simplify the process.