To solve the integral $\int \frac{x}{1 + x^4} \, dx$, we can proceed with a substitution method.

Step-by-Step Solution:

1. Substitution: Let $u = x^2$, so $du = 2x \, dx$, or equivalently $\frac{du}{2} = x \, dx$.

2. Rewrite the integral: Substitute $x^2 = u$, which means $x \, dx = \frac{du}{2}$. The integral becomes: $\int \frac{x}{1 + x^4} \, dx = \int \frac{\frac{1}{2} \, du}{1 + u^2}.$

3. Simplify: The integral now reduces to: $\frac{1}{2} \int \frac{1}{1 + u^2} \, du.$

4. Evaluate the integral: Recall that $\int \frac{1}{1 + u^2} \, du = \arctan(u) + C$. Thus: $\frac{1}{2} \int \frac{1}{1 + u^2} \, du = \frac{1}{2} \arctan(u) + C.$

5. Back-substitute $u = x^2$: Replace $u$ with $x^2$: $\frac{1}{2} \arctan(u) + C = \frac{1}{2} \arctan(x^2) + C.$

Final Answer:

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

Would you like further clarifications or additional examples?

Related Questions:

1. How is substitution used in other integrals involving $x^n$?
2. What is the derivative of $\arctan(x)$, and how does it relate to integrals?
3. Could partial fraction decomposition apply to this integral?
4. How do we integrate expressions of the form $\int \frac{x}{a + x^4} \, dx$?
5. What happens if the denominator is $1 - x^4$ instead of $1 + x^4$?

Tip:

Always check if substitution or trigonometric identities can simplify the problem before jumping to more advanced techniques.