int (x+5)/sqrt(9-(x-3)^2) dx Find the indefinite integral

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