Question

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

• A. 1
• B. 2
• C. -1
• D. -2

Answer 1

The correct option is A

1

$\int \frac{dx}{(1+\sqrt{x}{)}^{2010}}$

$=\int \frac{\sqrt{x}}{\sqrt{x}(1+\sqrt{x}{)}^{2010}}dx$

$=2\int \frac{t-1}{t^{2010}}dt$ $(t=1+\sqrt{x})$

$=2[\frac{1}{2009t^{2009}}-\frac{1}{2008t^{2008}}]+c$

$\Rightarrow \alpha =2009,\beta =2008$