Question

If $\int \frac{dx}{{(1+\sqrt{x})}^{2010}}=\frac{2}{\alpha {(1+\sqrt{x})}^{\alpha }}-\frac{2}{\beta {(1+\sqrt{x})}^{\beta }}+C,$ where $C$ is arbitrary constant of integration and $\alpha ,\beta >0,$ then

• A. $\alpha =2009,\beta =2008$
• B. $\alpha =2008,\beta =2009$
• C. $\alpha =2010,\beta =2008$
• D. $\alpha =2009,\beta =2009$

Answer 1

The correct option is A $\alpha =2009,\beta =2008$

Given : $\int \frac{dx}{(1+\sqrt{x}{)}^{2010}}$
Let $I=\int \frac{2\sqrt{x}dx}{2\sqrt{x}(1+\sqrt{x}{)}^{2010}}$
Put $1+\sqrt{x}=t\Rightarrow \frac{1}{2\sqrt{x}}dx=dt$
$I=2\int \frac{(t-1)dt}{t^{2010}}$
$=2\int (t^{-2009}-t^{-2010})dt$
$=2(\frac{t^{-2008}}{-2008}-\frac{t^{-2009}}{-2009})+C$
$I=\frac{2}{2009(1+\sqrt{x}{)}^{2009}}-\frac{2}{2008(1+\sqrt{x}{)}^{2008}}+C$
$\therefore \alpha =2009$ and $\beta =2008$