Evaluate: `int(sqrt(x)dx)/(1+x)`

Explanation

To evaluate the integral

\[ \int \frac{\sqrt{x}}{1+x} \, dx, \]

we can use the substitution method. Let's go through the solution step by step.

### Step 1: Substitution
Let \( x = t^2 \). Then, we have:

\[ dx = 2t \, dt. \]

### Step 2: Rewrite the Integral
Substituting \( x = t^2 \) into the integral gives:

\[ \sqrt{x} = \sqrt{t^2} = t, \]
and
\[ 1 + x = 1 + t^2. \]

Thus, the integral becomes:

\[ \int \frac{\sqrt{x}}{1+x} \, dx = \int \frac{t}{1+t^2} \cdot 2t \, dt = 2 \int \frac{t^2}{1+t^2} \, dt. \]

### Step 3: Simplify the Integral
We can simplify \( \frac{t^2}{1+t^2} \) as follows:

\[ \frac{t^2}{1+t^2} = 1 - \frac{1}{1+t^2}. \]

Thus, we can rewrite the integral:

\[ 2 \int \frac{t^2}{1+t^2} \, dt = 2 \int \left(1 - \frac{1}{1+t^2}\right) dt = 2 \int 1 \, dt - 2 \int \frac{1}{1+t^2} \, dt. \]

### Step 4: Evaluate the Integrals
Now, we evaluate each integral separately:

1. The first integral:

\[ 2 \int 1 \, dt = 2t. \]

2. The second integral:

\[ -2 \int \frac{1}{1+t^2} \, dt = -2 \tan^{-1}(t). \]

### Step 5: Combine the Results
Combining both results, we have:

\[ \int \frac{\sqrt{x}}{1+x} \, dx = 2t - 2 \tan^{-1}(t) + C. \]

### Step 6: Substitute Back
Now, substituting back \( t = \sqrt{x} \):

\[ = 2\sqrt{x} - 2 \tan^{-1}(\sqrt{x}) + C. \]

### Final Answer
Thus, the final result is:

\[ \int \frac{\sqrt{x}}{1+x} \, dx = 2\sqrt{x} - 2 \tan^{-1}(\sqrt{x}) + C. \]

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