This integral is:

$\int_0^{\sqrt{3}} \frac{4x}{\sqrt{x^2 + 1}} \, dx$

Solution:

To evaluate this integral, follow these steps:

1. Substitution: Let: $u = x^2 + 1$ Then: $du = 2x \, dx$

   Replace $x^2 + 1$ with $u$ and adjust for the constants: $2x \, dx = du \quad \implies \quad 4x \, dx = 2 \, du$

2. Adjust Limits: When $x = 0$, $u = 0^2 + 1 = 1$.
   When $x = \sqrt{3}$, $u = (\sqrt{3})^2 + 1 = 4$.

3. Rewrite the Integral: Substituting everything, the integral becomes: $\int_1^4 \frac{2}{\sqrt{u}} \, du$

   Simplify: $2 \int_1^4 u^{-1/2} \, du$

4. Integrate: Use the power rule: \int u^n \, du = \frac{u^{n+1}}{n+1} \quad \text{(for n \neq -1)}.

   Here, $n = -1/2$: $2 \int u^{-1/2} \, du = 2 \cdot \left( 2u^{1/2} \right) = 4u^{1/2}.$

5. Apply Limits: Substitute the limits $u = 1$ to $u = 4$: $4u^{1/2} \Big|_1^4 = 4 \left( \sqrt{4} - \sqrt{1} \right) = 4 \left( 2 - 1 \right) = 4.$

Final Answer:

$\boxed{4}$

Let me know if you'd like a deeper explanation for any step or have more questions!

Expanding Questions:

1. How does substitution simplify the evaluation of definite integrals?
2. Why is the power rule for integration important in this solution?
3. What is the role of adjusting the limits in substitution-based integration?
4. How would this integral differ if the numerator were a constant instead of $4x$?
5. Could this integral be solved without substitution? Why or why not?

Tip:

Always adjust the limits when changing variables in a definite integral—it avoids the need to back-substitute.