`int(dx)/((1+sqrtx)^(2010))=2[(1)/(alpha(1+sqrtx)^(alpha))-(1)/(beta(1+sqrtx))^(beta)]+c` where `alpha, betagt0" then "alpha-beta` is

Options

A. 1
B. 2
C. `-1`
D. `-2`

Explanation

To solve the integral

\[ \int \frac{dx}{(1 + \sqrt{x})^{2010}} = 2\left[\frac{1}{\alpha(1 + \sqrt{x})^{\alpha}} - \frac{1}{\beta(1 + \sqrt{x})^{\beta}}\right] + c \]

where \(\alpha, \beta > 0\), we need to find the value of \(\alpha - \beta\).

### Step-by-step Solution:

1. **Substitution**: Let \( t = 1 + \sqrt{x} \). Then, we have \(\sqrt{x} = t - 1\) and \(x = (t - 1)^2\).

2. **Differentiation**: Differentiate \(t\) with respect to \(x\):
\[ dt = \frac{1}{2\sqrt{x}} dx \implies dx = 2\sqrt{x} dt = 2(t - 1) dt \]

3. **Rewrite the Integral**: Substitute \(dx\) and \(t\) into the integral:
\[ \int \frac{dx}{(1 + \sqrt{x})^{2010}} = \int \frac{2(t - 1) dt}{t^{2010}} \]

4. **Separate the Integral**: This can be separated into two integrals:
\[ = 2 \int \left( \frac{t}{t^{2010}} - \frac{1}{t^{2010}} \right) dt = 2 \int \left( t^{-2009} - t^{-2010} \right) dt \]

5. **Integrate**: Now, integrate each term:
\[ = 2 \left( \frac{t^{-2008}}{-2008} - \frac{t^{-2009}}{-2009} \right) + C \]
\[ = -\frac{2}{2008} t^{-2008} + \frac{2}{2009} t^{-2009} + C \]

6. **Substitute Back**: Substitute \(t = 1 + \sqrt{x}\) back into the equation:
\[ = -\frac{2}{2008(1 + \sqrt{x})^{2008}} + \frac{2}{2009(1 + \sqrt{x})^{2009}} + C \]

7. **Identify \(\alpha\) and \(\beta\)**: From the equation, we can compare:
- \(\alpha = 2009\)
- \(\beta = 2008\)

8. **Calculate \(\alpha - \beta\)**:
\[ \alpha - \beta = 2009 - 2008 = 1 \]

### Final Answer:
\[ \alpha - \beta = 1 \]