We are solving the integral:

$I = \int \frac{x}{x^2 + 4} \, dx$

Step 1: Substitution

Let: $u = x^2 + 4 \quad \text{so that} \quad du = 2x \, dx.$ Rewriting $x \, dx$ in terms of $u$: $x \, dx = \frac{1}{2} \, du.$

Substitute $u$ into the integral: $I = \int \frac{1}{u} \cdot \frac{1}{2} \, du.$

Step 2: Simplify and Integrate

The integral becomes: $I = \frac{1}{2} \int \frac{1}{u} \, du.$

The integral of $\frac{1}{u}$ is $\ln |u|$, so: $I = \frac{1}{2} \ln |u| + C.$

Step 3: Back-Substitute $u$

Recall $u = x^2 + 4$. Substituting back: $I = \frac{1}{2} \ln |x^2 + 4| + C.$

Since $x^2 + 4 > 0$ for all real $x$, the absolute value can be omitted: $I = \frac{1}{2} \ln(x^2 + 4) + C.$

Final Answer:

$\boxed{\int \frac{x}{x^2 + 4} \, dx = \frac{1}{2} \ln(x^2 + 4) + C.}$

Let me know if you'd like further clarification or details!

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Here are 5 related questions to explore:

1. How do substitutions simplify integrals?
2. Can you explain the integral of $\frac{1}{x^2 + a^2}$?
3. How would this integral change if the numerator was $x^2$?
4. How does the logarithmic function arise in integration?
5. What are the properties of definite integrals involving logarithms?

Tip: For substitutions, always ensure your $u$-substitution simplifies both the numerator and denominator effectively.