To integrate \( \int \frac{\sqrt{x}}{x+1} \, dx \), let's proceed with a substitution:

1. Set \( u = \sqrt{x} \), so \( x = u^2 \).
2. Then, \( dx = 2u \, du \).

Substituting these into the integral:

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

Now, rewrite the numerator \( 2u^2 \) as \( 2(u^2 + 1 - 1) = 2(u^2 + 1) - 2 \):

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

This separates into two integrals:

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

Now, integrate each term separately:

1. For \( \int 2 \, du = 2u \).
2. For \( \int \frac{2}{u^2 + 1} \, du = 2 \arctan(u) \).

Combining these results:

\[
\int \frac{\sqrt{x}}{x+1} \, dx = 2u - 2 \arctan(u) + C
\]

Substitute \( u = \sqrt{x} \) back:

\[
= 2\sqrt{x} - 2 \arctan(\sqrt{x}) + C
\]

So the final answer is:

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

Would you like more details or have any questions about this solution?

Here are related questions to explore:

1. How does the substitution method work in integration?
2. Why is it beneficial to rewrite the integrand in terms of simpler expressions?
3. How does the arctan function arise in integration, and what does it represent?
4. How does integration by parts differ from the substitution method?
5. How would this problem change if we were integrating \(\frac{1}{\sqrt{x}(x+1)}\)?

**Tip:** In integration, rewriting the integrand can simplify the process, especially when using substitution.

Integration of \( \frac{\sqrt{x}}{x+1} \, dx \)

Solve the integral \( \int \frac{\sqrt{x}}{x+1} \, dx \) with a step-by-step solution using substitution. Learn to rewrite integrals with trigonometric functions like \( \arctan(u) \), applying key integration techniques suitable for advanced high school and college-level calculus students.

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