Evaluate :
(ii) `int_(0)^(1) (x)/(1+sqrt(x))dx`

Answer

To evaluate the integral \( \int_{0}^{1} \frac{x}{1+\sqrt{x}} \, dx \), we can use substitution. Let's follow the steps to solve this integral.

### Step 1: Substitution
Let \( t = \sqrt{x} \). Then, we have:
\[ x = t^2 \]
Differentiating both sides gives:
\[ dx = 2t \, dt \]

### Step 2: Change the limits
When \( x = 0 \):
\[ t = \sqrt{0} = 0 \]
When \( x = 1 \):
\[ t = \sqrt{1} = 1 \]
So, the limits of integration change from \( x = 0 \) to \( x = 1 \) to \( t = 0 \) to \( t = 1 \).

### Step 3: Substitute in the integral
Now substitute \( x \) and \( dx \) in the integral:
\[ \int_{0}^{1} \frac{x}{1+\sqrt{x}} \, dx = \int_{0}^{1} \frac{t^2}{1+t} \cdot 2t \, dt \]
This simplifies to:
\[ = 2 \int_{0}^{1} \frac{t^3}{1+t} \, dt \]

### Step 4: Simplifying the integral
Now we can simplify the integrand:
\[ \frac{t^3}{1+t} = t^3 \cdot \frac{1}{1+t} \]
We can split this into two parts:
\[ = t^3 - \frac{t^3}{1+t} \]
Thus, we rewrite the integral:
\[ = 2 \left( \int_{0}^{1} t^3 \, dt - \int_{0}^{1} \frac{t^3}{1+t} \, dt \right) \]

### Step 5: Evaluate the first integral
The first integral is straightforward:
\[ \int_{0}^{1} t^3 \, dt = \left[ \frac{t^4}{4} \right]_{0}^{1} = \frac{1}{4} \]

### Step 6: Evaluate the second integral
Now we need to evaluate \( \int_{0}^{1} \frac{t^3}{1+t} \, dt \). We can use polynomial long division or a series expansion, but here we can also use the formula for the integral:
\[ \int \frac{t^n}{1+t} \, dt = t^n - \int \frac{t^{n-1}}{1+t} \, dt \]
For \( n = 3 \):
\[ \int_{0}^{1} \frac{t^3}{1+t} \, dt = \int_{0}^{1} t^3 \, dt - \int_{0}^{1} \frac{t^2}{1+t} \, dt \]
We already calculated \( \int_{0}^{1} t^3 \, dt = \frac{1}{4} \).

### Step 7: Evaluate \( \int_{0}^{1} \frac{t^2}{1+t} \, dt \)
Using the same method:
\[ \int_{0}^{1} \frac{t^2}{1+t} \, dt = \int_{0}^{1} t^2 \, dt - \int_{0}^{1} \frac{t}{1+t} \, dt \]
Calculating \( \int_{0}^{1} t^2 \, dt = \frac{1}{3} \).

### Step 8: Combine results
Now we can combine our results:
\[ \int_{0}^{1} \frac{t^3}{1+t} \, dt = \frac{1}{4} - \left( \frac{1}{3} - \int_{0}^{1} \frac{t}{1+t} \, dt \right) \]
The integral \( \int_{0}^{1} \frac{t}{1+t} \, dt \) can be calculated as:
\[ = \int_{0}^{1} (1 - \frac{1}{1+t}) \, dt = 1 - \log(2) \]

### Final Calculation
Putting it all together:
\[ 2 \left( \frac{1}{4} - \left( \frac{1}{3} - (1 - \log(2)) \right) \right) \]
This simplifies to:
\[ = 2 \left( \frac{1}{4} - \frac{1}{3} + 1 - \log(2) \right) \]
Calculating this gives:
\[ = 2 \left( \frac{3}{12} - \frac{4}{12} + \frac{12}{12} - \log(2) \right) = 2 \left( \frac{11}{12} - \log(2) \right) = \frac{22}{12} - 2\log(2) = \frac{11}{6} - 2\log(2) \]

Thus, the final answer is:
\[ \frac{11}{6} - 2\log(2) \]