To integrate the function:

$\int \frac{x}{2x^2 + 1} \, dx$

We can solve this using substitution. Let's follow these steps:

Step 1: Choose a substitution

Let $u = 2x^2 + 1$. Then, differentiate both sides with respect to $x$:

$\frac{du}{dx} = 4x$

or equivalently,

$du = 4x \, dx$

Now, solve for $x \, dx$:

$x \, dx = \frac{du}{4}$

Step 2: Substitute into the integral

Now substitute $u$ and $x \, dx$ into the original integral:

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

Simplifying:

$\int \frac{1}{u} \cdot \frac{du}{4} = \frac{1}{4} \int \frac{1}{u} \, du$

Step 3: Integrate

The integral of $\frac{1}{u}$ is $\ln|u|$, so we have:

$\frac{1}{4} \ln|u| + C$

Step 4: Substitute back $u = 2x^2 + 1$

Now substitute $u$ back into the expression:

$\frac{1}{4} \ln(2x^2 + 1) + C$

Final Answer:

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

---

Would you like a more detailed explanation of any part of this process? Here are some related questions that might interest you:

1. How do you handle integrals that involve a quadratic expression in the denominator?
2. What other substitution techniques are helpful for solving integrals?
3. Can this method be extended to rational functions with higher powers?
4. How can we solve integrals of the form $\int \frac{dx}{ax^2 + bx + c}$?
5. How does the logarithmic rule $\int \frac{1}{u} du = \ln|u| + C$ apply in more complex cases?

Tip: Always look for a substitution that simplifies the integral. In this case, identifying the quadratic expression in the denominator was key to applying the substitution effectively.