Evaluate the following integral:

re-writing the given equation as

$\int \frac{(x-3)+3}{(x-3) \sqrt{x+1}} d x$

$\int \frac{(x-3)}{(x-3) \sqrt{x+1}} d x+\int \frac{3}{(x-3) \sqrt{x+1}} d x$

For the first part using identity $\int x^{n} d x=\frac{x^{n+1}}{n+1}$

$2 \sqrt{x+1}+c$

For the second part

assume $x+1=t^{2}$

$\mathrm{d} x=2 \mathrm{tdt}$

$\int \frac{2 d t}{\left(t^{2}-4\right)}$

Using identity $\int \frac{d z}{(z)^{2}-1}=\frac{1}{2} \log \left|\frac{z-1}{z+1}\right|+c$

$\frac{1}{2} \log \left|\frac{\mathrm{t}-2}{\mathrm{t}+2}\right|+\mathrm{c}$

$\frac{1}{2} \log \left|\frac{\sqrt{(\mathrm{x}+2)}-2}{\sqrt{\mathrm{x}+2}+2}\right|+\mathrm{c}$

Hence integral is

$\frac{1}{2} \log \left|\frac{\sqrt{(x+2)}-2}{\sqrt{x+2}+2}\right|+c+2 \sqrt{x+1}$