To solve the indefinite integral \(\int \frac{x+9}{\sqrt{16-(x-4)^{2}}} dx\), we can use a trigonometric substitution. Recognize that the expression under the square root is of the form \(a^2 - u^2\), which suggests a substitution of the form \(u = a \sin(\theta)\). Here, \(a = 4\) and \(u = x - 4\), so we can let \(x - 4 = 4 \sin(\theta)\). This substitution will simplify the integral into a form that can be integrated using standard trigonometric identities.
We start with the integral
\[
\int \frac{x+9}{\sqrt{16-(x-4)^{2}}} dx.
\]
Using the substitution \(u = x - 4\), we have \(x = u + 4\). The integral becomes
\[
\int \frac{u + 13}{\sqrt{16 - u^2}} du.
\]
Next, we apply the trigonometric substitution \(u = 4 \sin(\theta)\), which gives \(du = 4 \cos(\theta) d\theta\). The integral transforms to
\[
\int \frac{4 \sin(\theta) + 13}{\sqrt{16 - 16 \sin^2(\theta)}} \cdot 4 \cos(\theta) d\theta.
\]
The expression under the square root simplifies as follows:
\[
\sqrt{16 - 16 \sin^2(\theta)} = \sqrt{16(1 - \sin^2(\theta))} = 4 \cos(\theta).
\]
Thus, the integral simplifies to
\[
\int \frac{(4 \sin(\theta) + 13) \cdot 4 \cos(\theta)}{4 \cos(\theta)} d\theta = \int (4 \sin(\theta) + 13) d\theta.
\]
Now we can integrate:
\[
\int (4 \sin(\theta) + 13) d\theta = -4 \cos(\theta) + 13\theta + C.
\]
We need to revert back to the variable \(x\). From our substitution \(u = 4 \sin(\theta)\), we have
\[
\sin(\theta) = \frac{u}{4} = \frac{x - 4}{4}.
\]
Thus,
\[
\cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - \left(\frac{x - 4}{4}\right)^2} = \frac{\sqrt{16 - (x - 4)^2}}{4}.
\]
Also,
\[
\theta = \arcsin\left(\frac{x - 4}{4}\right).
\]
Substituting back, we get:
\[
-4 \cdot \frac{\sqrt{16 - (x - 4)^2}}{4} + 13 \arcsin\left(\frac{x - 4}{4}\right) + C = -\sqrt{16 - (x - 4)^2} + 13 \arcsin\left(\frac{x - 4}{4}\right) + C.
\]
Thus, the indefinite integral is
\[
\boxed{-\sqrt{16 - (x - 4)^2} + 13 \arcsin\left(\frac{x - 4}{4}\right) + C}.
\]
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