The given integral to solve is:

$\int_{0}^{1} \frac{x}{2x^2 + 1} \, dx$

Solution:

Let us perform the substitution method:

1. Substitute $u = 2x^2 + 1$:

   • Then, $du = 4x \, dx$, so $\frac{du}{4} = x \, dx$.

2. Change the limits of integration according to $u = 2x^2 + 1$:

   • When $x = 0$, $u = 2(0)^2 + 1 = 1$,
   • When $x = 1$, $u = 2(1)^2 + 1 = 3$.

Thus, the integral becomes: $\int_{1}^{3} \frac{1}{u} \cdot \frac{1}{4} \, du = \frac{1}{4} \int_{1}^{3} \frac{1}{u} \, du$

3. Evaluate the integral: $\frac{1}{4} \int_{1}^{3} \frac{1}{u} \, du = \frac{1}{4} \left[ \ln|u| \right]_{1}^{3}$

4. Apply the limits: $\frac{1}{4} \left[ \ln(3) - \ln(1) \right] = \frac{1}{4} \ln(3)$

5. Final Answer: $\boxed{\frac{\ln(3)}{4}}$

Let me know if you'd like a more detailed explanation or have any questions! 😊

Related Questions:

1. What is the substitution rule in integration, and why is it useful?
2. How do logarithmic properties simplify integration results?
3. Can this integral be solved without substitution?
4. What happens if we use a different substitution for this integral?
5. How can definite integrals be interpreted geometrically?

Tip:

Always check the limits of integration carefully when performing substitutions to avoid errors.