Calculus of a Single Variable, Chapter 5, 5.7, Section 5.7, Problem 19

We have to evaluate the integral \int \frac{x+5}{\sqrt{9-(x-3)^2}}dx

Let x-3=u
So, dx=du
Hence we can write,
\int \frac{x+5}{\sqrt{9-(x-3)^2}}dx=\int \frac{u+8}{\sqrt{9-u^2}}du
=\int \frac{u}{\sqrt{9-u^2}}du+\int \frac{8}{\sqrt{9-u^2}}dx
Now we will first evaluate the integral: \int \frac{udu}{\sqrt{9-u^2}}
Let 9-u^2=t
So, -2udu=dt
Hence we have,
\int \frac{udu}{\sqrt{9-u^2}}=\int \frac{-dt}{2\sqrt{t}}
=-\sqrt{t}
=-\sqrt{9-u^2}

Now we will evaluate the integral\int \frac{8}{\sqrt{9-u^2}}du

\int \frac{8}{\sqrt{9-u^2}}du=\frac{8du}{3\sqrt{1-(\frac{u}{3})^2}}
=8sin^{-1}(\frac{u}{3}) since we have the identity
\frac{d}{dx}(sin^{-1}(\frac{u}{a}))=\frac{1}{a}.\frac{1}{\sqrt{1-(\frac{u}{a})^2}}

So finally we have the result as:
\int \frac{x+5}{\sqrt{9-(x-3)^2}}dx=-\sqrt{9-u^2}+8sin^{-1}(\frac{u}{3})+C
=-\sqrt{9-(x-3)^2}+8sin^{-1}(\frac{x-3}{3})+C