How can i get the answer to this improper integral: $\int_{0}^{1}\frac{1}{(x-2)\sqrt{x^2+2x} } \;\mathrm{d}x $ ?

You may use the limit comparison test. Note that at $x=0$, there is division by zero, so this is an improper integral. Near $x=0$ we have 
\[\frac{1}{(x-2)\sqrt{x^2+2x}}\approx \frac{1}{(0-2)\sqrt{0^2+2x}}\approx -\frac{1}{2\sqrt{2}\sqrt{x}}.\]
This means that 
\[\lim_{x \rightarrow 0} \frac{\frac{1}{(x-2)\sqrt{x^2+2x}}}{-\frac{1}{2\sqrt{2}\sqrt{x}}}=1.\]
It is easy ti see that the improper integral 
\[\int_0^1 -\frac{1}{2\sqrt{2}\sqrt{x}}dx\]
is convergent, and hence by the Limit Comparison Theorem $\int_0^1 \frac{1}{(x-2)\sqrt{x^2+2x}} dx$ also converges, and since on $(0,1)$ we have 
\[\frac{1}{(x-2)\sqrt{x^2+2x}}<0,\]
this improper integral should converge to a negative number.