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 a constant of integration and $\alpha ,\beta >0,$ then

• A. $|\alpha -\beta |=1$
• B. $\frac{\beta +2}{(2010{)}^2}=\frac{1}{\alpha +1}$
• C. $\beta ,\alpha ,2010$ (taken in that order) are in arithmetic progression.
• D. $\alpha +1=\beta +1=2010$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $|\alpha -\beta |=1$
B $\frac{\beta +2}{(2010{)}^2}=\frac{1}{\alpha +1}$
C $\beta ,\alpha ,2010$ (taken in that order) are in arithmetic progression.

$I=\int \frac{2\sqrt{x}dx}{2\sqrt{x}(1+\sqrt{x}{)}^{2010}}$
Put $1+\sqrt{x}=t\Rightarrow \frac{1}{2\sqrt{x}}dx=dt$
$I=2\int \frac{(t-1)dt}{t^{2010}}$
$=2\int (t^{-2009}-t^{-2010})dt$

$=2(\frac{t^{-2008}}{-2008}-\frac{t^{-2009}}{-2009})+C$
$=\frac{2}{2009(1+\sqrt{x}{)}^{2009}}-\frac{2}{2008(1+\sqrt{x}{)}^{2008}}+C$
$\therefore \alpha =2009,\beta =2008$.