`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`

                          ``