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.