To integrate $\frac{2x}{dx}$, it appears that you meant $\int 2x \, dx$, which is a straightforward integral.

Solution:

We apply the basic power rule for integration:

\int x^n \, dx = \frac{x^{n+1}}{n+1} + C, \quad \text{(for n \neq -1)}.

Here, $2x$ is equivalent to $2 \cdot x^1$.

$\int 2x \, dx = 2 \cdot \frac{x^{1+1}}{1+1} + C = 2 \cdot \frac{x^2}{2} + C.$

Simplify:

$\int 2x \, dx = x^2 + C.$

Final Answer:

$x^2 + C$

Would you like further explanation on integration rules or examples?

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Related Questions:

1. How do you integrate a function like $\int 3x^2 \, dx$?
2. What is the rule for integrating $x^n$ when $n = -1$?
3. How do you find the definite integral of $\int_{0}^{2} 2x \, dx$?
4. How does the constant of integration $C$ impact solutions to indefinite integrals?
5. What are some real-world applications of integrals involving $x$-functions?

Tip:

Always double-check if the integral is definite (bounds are given) or indefinite (no bounds, include $C$).