$$ f ^ { \prime } ( x ) = e ^ { 2 x } [ 2 \ln x + \frac { 1 } { x } ] $$

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We want to find the antiderivative of  
$$
f'(x)=e^{2x}\left(2\ln x+\frac{1}{x}\right)
$$
That is, we need to find $$ f(x) $$ such that  
$$
f(x)=\int e^{2x}\left(2\ln x+\frac{1}{x}\right)\,dx.
$$

**Step 1. Look for a candidate whose derivative matches $$ f'(x) $$**

Consider the function  
$$
g(x)=e^{2x}\ln x.
$$
We compute the derivative $$ g'(x) $$ using the product rule:
$$
\frac{d}{dx}\left(e^{2x}\ln x\right)=\frac{d}{dx}\left(e^{2x}\right)\ln x+e^{2x}\frac{d}{dx}\left(\ln x\right).
$$

**Step 2. Compute the derivative components**

- The derivative of $$e^{2x}$$ is:
  $$
  \frac{d}{dx}\left(e^{2x}\right)=2e^{2x}.
  $$
  
- The derivative of $$\ln x$$ is:
  $$
  \frac{d}{dx}\left(\ln x\right)=\frac{1}{x}.
  $$

**Step 3. Substitute into the product rule**

Using these results, we have:
$$
g'(x)=2e^{2x}\ln x+e^{2x}\frac{1}{x}.
$$

Which can be written as:
$$
g'(x)=e^{2x}\left(2\ln x+\frac{1}{x}\right).
$$

**Step 4. Conclusion**

Notice that $$ g'(x) $$ is exactly the same as $$ f'(x) $$. Therefore, we conclude:
$$
f(x)=e^{2x}\ln x+C,
$$
where $$ C $$ is an arbitrary constant of integration.
The antiderivative of $$ f'(x) = e^{2x}\left(2\ln x + \frac{1}{x}\right) $$ is $$ f(x) = e^{2x}\ln x + C $$, where $$ C $$ is a constant.

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