Evaluate: `int_0^1 1/(sqrt(1+x)-sqrtx) dx`

Answer

To evaluate the integral \( I = \int_0^1 \frac{1}{\sqrt{1+x} - \sqrt{x}} \, dx \), we will follow a systematic approach.

### Step 1: Rationalize the Denominator
We start by rationalizing the denominator. We multiply and divide by the conjugate of the denominator:

\[ I = \int_0^1 \frac{\sqrt{1+x} + \sqrt{x}}{(\sqrt{1+x} - \sqrt{x})(\sqrt{1+x} + \sqrt{x})} \, dx \]

This simplifies to:

\[ I = \int_0^1 \frac{\sqrt{1+x} + \sqrt{x}}{1+x - x} \, dx = \int_0^1 \frac{\sqrt{1+x} + \sqrt{x}}{1} \, dx = \int_0^1 (\sqrt{1+x} + \sqrt{x}) \, dx \]

### Step 2: Split the Integral
Now we can split the integral into two parts:

\[ I = \int_0^1 \sqrt{1+x} \, dx + \int_0^1 \sqrt{x} \, dx \]

### Step 3: Evaluate Each Integral
1. **Evaluate \( \int_0^1 \sqrt{x} \, dx \)**:
\[ \int_0^1 \sqrt{x} \, dx = \int_0^1 x^{1/2} \, dx = \left[ \frac{x^{3/2}}{3/2} \right]_0^1 = \left[ \frac{2}{3} x^{3/2} \right]_0^1 = \frac{2}{3} \]

2. **Evaluate \( \int_0^1 \sqrt{1+x} \, dx \)**:
To evaluate this integral, we can use the substitution \( u = 1+x \), hence \( du = dx \) and the limits change from \( x=0 \) to \( u=1 \) and \( x=1 \) to \( u=2 \):
\[ \int_0^1 \sqrt{1+x} \, dx = \int_1^2 \sqrt{u} \, du = \left[ \frac{u^{3/2}}{3/2} \right]_1^2 = \left[ \frac{2}{3} u^{3/2} \right]_1^2 = \frac{2}{3} \left( 2^{3/2} - 1^{3/2} \right) = \frac{2}{3} \left( 2\sqrt{2} - 1 \right) \]

### Step 4: Combine the Results
Now we combine the results of the two integrals:

\[ I = \frac{2}{3} \left( 2\sqrt{2} - 1 \right) + \frac{2}{3} \]

This simplifies to:

\[ I = \frac{2}{3} (2\sqrt{2}) = \frac{4\sqrt{2}}{3} \]

### Final Result
Thus, the value of the integral is:

\[ \boxed{\frac{4\sqrt{2}}{3}} \]