We are tasked with evaluating the integral:

\[
I = \int \frac{x}{\sqrt{1 + x^4}} \, dx
\]

### Step 1: Substitution
A good approach for integrals involving expressions like \(1 + x^4\) is to use a substitution. Let’s set:

\[
u = x^2
\]

Then, the derivative of \(u\) with respect to \(x\) is:

\[
du = 2x \, dx \quad \Rightarrow \quad \frac{du}{2} = x \, dx
\]

### Step 2: Rewrite the Integral
Now, substitute this into the original integral:

\[
I = \int \frac{x}{\sqrt{1 + x^4}} \, dx = \int \frac{1}{\sqrt{1 + u^2}} \cdot \frac{du}{2}
\]

Thus, the integral becomes:

\[
I = \frac{1}{2} \int \frac{du}{\sqrt{1 + u^2}}
\]

### Step 3: Use a Standard Integral
The integral:

\[
\int \frac{du}{\sqrt{1 + u^2}}
\]

is a standard form, whose result is:

\[
\text{arsinh}(u) = \ln \left( u + \sqrt{1 + u^2} \right)
\]

### Step 4: Back-Substitute
Substitute back \(u = x^2\) to get:

\[
I = \frac{1}{2} \ln \left( x^2 + \sqrt{1 + x^4} \right) + C
\]

### Final Answer:
The evaluated integral is:

\[
I = \frac{1}{2} \ln \left( x^2 + \sqrt{1 + x^4} \right) + C
\]

---

Would you like further details or have any questions?

Here are 5 related questions:
1. How do you recognize when substitution is the best method?
2. What is the difference between using trigonometric and hyperbolic substitutions?
3. Can you explain the concept of the arcsinh function in detail?
4. How would the integral change if we had \(1 - x^4\) instead of \(1 + x^4\)?
5. What other methods could we use to solve similar integrals?

**Tip**: Always check if your integral matches a standard form—this can save you significant time!

Learn how to solve the integral \int \frac{x}{\sqrt{1+x^4}} dx using substitution and hyperbolic functions. This step-by-step solution explains key calculus concepts, including the substitution method and the use of standard integrals. Suitable for university-level calculus students.

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