91Ó°ÊÓ

The integral of a sum (or difference) of functions is equal to the sum (or difference) of their integrals. We can rewrite: \[\int \frac{x+5}{\sqrt{9-(x-3)^{2}}} dx \]as\[\int \frac{x}{\sqrt{9-(x-3)^{2}}} dx + \int \frac{5}{\sqrt{9-(x-3)^{2}}} dx\]

To tackle the first integral, notice that its structure suggests a substitution. Let's put \(u = x - 3\). We get the differential \(du = dx\) and our integral becomes \( \int \frac{u+3}{\sqrt{9-u^2}} du \) which splits into \( 3 \int \frac{du}{\sqrt{9-u^2}} + \int \frac{u}{\sqrt{9-u^2}} du \).The first part of the integrated function now resembles the integral of \( \frac{du}{\sqrt{a^2-u^2}} \), which is \( \arcsin(\frac{u}{a}) \). Thus, we can immediately evaluate this as \( 3 \arcsin (\frac{u}{3}) \).The second part is an example of a function whose derivative is in the numerator. The integral of \( \frac{u}{\sqrt{a^2 - u^2}} \) is \( -\sqrt{a^2 - u^2} \). Thus, this second integral can be evaluated as \( -\sqrt{9 - u^2} = -\sqrt{9 - (x-3)^2} \).

Now, for the second integral, we can directly write down the antiderivative because it matches the form \( \int \frac{du}{\sqrt{a^2-u^2}} \), which we know integrates to \( \arcsin(\frac{u}{a}) \). In our case, a = 3 and u = x - 3. So, this integral gives us \( 5 \arcsin (\frac{x-3}{3}) \).

Sum the results of the two integrations. Thus, our final expression for the integral is:\[3 \arcsin (\frac{x-3}{3}) -\sqrt{9 - (x-3)^2} + 5 \arcsin (\frac{x-3}{3}) + C\]Simplify this to:\[8 \arcsin (\frac{x-3}{3}) - \sqrt{9 - (x-3)^2} + C\]where C denotes the constant of integration.