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The integral you provided is:

\[
\int_1^{\sqrt[4]{5}} \left( \frac{\sqrt[3]{x^5}}{\sqrt[3]{x^2}} + 2x \right) dx
\]

Let me break it down and solve this step-by-step:

### Step 1: Simplifying the Expression
We have two terms in the integrand:
\[
\frac{\sqrt[3]{x^5}}{\sqrt[3]{x^2}} + 2x
\]

The first term simplifies as follows:
\[
\frac{\sqrt[3]{x^5}}{\sqrt[3]{x^2}} = \sqrt[3]{x^5} \cdot \frac{1}{\sqrt[3]{x^2}} = \sqrt[3]{x^{5-2}} = \sqrt[3]{x^3} = x
\]

So, the entire integrand becomes:
\[
x + 2x = 3x
\]

### Step 2: Performing the Integral
Now, we need to compute the integral:
\[
\int_1^{\sqrt[4]{5}} 3x \, dx
\]

The antiderivative of \(3x\) is:
\[
\frac{3x^2}{2}
\]

### Step 3: Applying the Limits of Integration
Now, evaluate the definite integral:
\[
\left[ \frac{3x^2}{2} \right]_1^{\sqrt[4]{5}} = \frac{3}{2} \left( (\sqrt[4]{5})^2 - 1^2 \right)
\]

Simplifying:
\[
(\sqrt[4]{5})^2 = \sqrt{5}, \quad 1^2 = 1
\]
\[
\frac{3}{2} (\sqrt{5} - 1)
\]

Thus, the result of the integral is:
\[
\frac{3}{2} (\sqrt{5} - 1)
\]

### Final Answer:
\[
\boxed{\frac{3}{2} (\sqrt{5} - 1)}
\]

Let me know if you would like further details or clarifications!

Here are five related questions to consider:
1. How would the integral change if the limits were different, say from 0 to 2?
2. What if the integrand had a different coefficient, like \( 4x \) instead of \( 3x \)?
3. How do you compute the integral if the integrand were \(x^2 + 2x\) instead?
4. Can you perform an indefinite integral of \( \frac{\sqrt[3]{x^5}}{\sqrt[3]{x^2}} + 2x \)?
5. How does the Fundamental Theorem of Calculus apply to this definite integral?

**Tip:** Simplifying expressions inside the integrand before integrating can often make the problem easier to solve.

Evaluate the definite integral \( \int_1^{\sqrt[4]{5}} \left( \frac{\sqrt[3]{x^5}}{\sqrt[3]{x^2}} + 2x \right) dx \).

A step-by-step solution for evaluating the definite integral \( \int_1^{\sqrt[4]{5}} \left( \frac{\sqrt[3]{x^5}}{\sqrt[3]{x^2}} + 2x \right) dx \). This problem involves simplifying radicals and applying the properties of exponents, suitable for students in Grades 11-12 or introductory calculus courses.

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