Evaluate the following integral:

Solution:

re-writing the given equation as

$\int \frac{(x-1)+2}{(x-1) \sqrt{x+2}} d x$

Now splitting the integral in two parts

$\int \frac{(x-1)}{(x-1) \sqrt{x+2}} d x+\int \frac{2}{(x-1) \sqrt{x+2}} d x$

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

$2 \sqrt{x+2}$

For the second part

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

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

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

$\int \frac{4 d t}{\left(t^{2}-3\right)}$

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

$\frac{2}{\sqrt{3}} \log \left|\frac{t-\sqrt{3}}{t+\sqrt{3}}\right|+c$

$\frac{2}{\sqrt{3}} \log \left|\frac{\sqrt{(x+2)}-\sqrt{3}}{\sqrt{x+2}+\sqrt{3}}\right|+c$

Hence integral is

$2 \sqrt{x+2}+\frac{2}{\sqrt{3}} \log \left|\frac{\sqrt{(x+2)}-\sqrt{3}}{\sqrt{x+2}+\sqrt{3}}\right|+c$