integration

\hspace{-16}$ Given $\int\frac{1}{(1+\sqrt{x})^{2010}}dx=2.\left\{\frac{1}{\alpha.(1+\ \sqrt{x})^{\alpha}}-\frac{1}{\beta.(1+\sqrt{x})^{\beta}}\right\}+C$\\\\\\ Now Calculating Value of $\int\frac{1}{(1+\sqrt{x})^{2010}}dx$\\\\\\ Put $(1+\sqrt{x})=t\Leftrightarrow \frac{1}{2.\sqrt{x}}dx=dt\Leftrightarrow dx=2\sqrt{x}dt=2.(t-1)dt$\\\\\\ $=\int\frac{2.(t-1)}{t^{2010}}dt=2.\left\{\int\frac{1}{t^{2009}}dt-\int\frac{1}{t^{2010}}dt\right\}$\\\\\\ $=2.\left\{-\frac{1}{2008}.\frac{1}{t^{2008}}+\frac{1}{2009}.\frac{1}{t^{2009}}\right\}+C$\\\\\\ $=2.\left\{-\frac{1}{2008}.\frac{1}{(1+\sqrt{x})^{2008}}+\frac{1}{2009}.\frac{1}{(1+\sqrt{x})^{2009}}\right\}+C$\\\\\\ So Here We get\\\\\\ $\int\frac{1}{(1+\sqrt{x})^{2010}}dx=2.\left\{\frac{1}{2009}.\frac{1}{(1+\sqrt{x})^{2009}}-\frac{1}{2008}.\frac{1}{(1+\sqrt{x})^{2008}}\right\}+C$\\\\\\ So From Here We Get $\boxed{\boxed{\mathbf{\alpha =2009}}}$ and $\boxed{\boxed{\mathbf{\beta =2008}}}$

I got that without solving it :)