We are tasked with finding the integral of:

\[
\int \frac{2\sqrt{x}}{1+x} \, dx
\]

### Step 1: Substitution
We'll perform a substitution to simplify this integral. Let:

\[
u = \sqrt{x} \quad \text{so that} \quad x = u^2 \quad \text{and} \quad dx = 2u \, du
\]

Now substitute these into the integral:

\[
\int \frac{2\sqrt{x}}{1+x} \, dx = \int \frac{2u}{1+u^2} \cdot 2u \, du = 4 \int \frac{u^2}{1+u^2} \, du
\]

### Step 2: Simplifying the integrand
We now rewrite the integrand:

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

Thus, the integral becomes:

\[
4 \int \left( 1 - \frac{1}{1+u^2} \right) du = 4 \left( \int 1 \, du - \int \frac{1}{1+u^2} \, du \right)
\]

### Step 3: Solving the integrals
- The integral of 1 with respect to \( u \) is simply \( u \).
- The integral of \( \frac{1}{1+u^2} \) is \( \tan^{-1}(u) \).

Thus, the integral is:

\[
4 \left( u - \tan^{-1}(u) \right) + C
\]

### Step 4: Substituting back \( u = \sqrt{x} \)
Finally, substitute \( u = \sqrt{x} \) back into the result:

\[
4 \left( \sqrt{x} - \tan^{-1}(\sqrt{x}) \right) + C
\]

### Final Answer:
The integral is:

\[
\boxed{4 \left( \sqrt{x} - \tan^{-1}(\sqrt{x}) \right) + C}
\]

Would you like further details on any step or have any questions?

Here are 5 related questions for practice:
1. What is the integral of \( \frac{\sqrt{x}}{1+x^2} \)?
2. How would you integrate \( \frac{x}{1+x^2} \)?
3. Can you solve \( \int \frac{2x}{1+x^2} \, dx \)?
4. What substitution can you use for \( \int \frac{\sqrt{x}}{x+2} \, dx \)?
5. How does the method of partial fractions apply to integrals of rational functions?

**Tip:** When dealing with square roots in integrals, substitutions like \( u = \sqrt{x} \) are often helpful in simplifying the expression.

Learn how to solve the integral of (2√x)/(1+x) step by step using substitution and trigonometric integrals. This advanced calculus problem is ideal for university-level students and involves techniques such as substitution with u = √x and recognizing the inverse tangent function.

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