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

Answer 1

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

$\beta ,\alpha ,2010$ are in A.P.

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

$|\alpha -\beta |=|2009-2008|=1$

$\frac{\beta +2}{(2010{)}^2}=\frac{1}{2010}$ and $\frac{1}{\alpha +1}=\frac{1}{2010}$
$\therefore \frac{\beta +2}{(2010{)}^2}=\frac{1}{\alpha +1}$

Clearly, $\beta ,\alpha ,2010$ are in A.P.

$\alpha +1=2010$
$\beta +1=2009$
$\therefore \alpha +1\ne \beta +1$