To solve the equation \(\sqrt{x+3} + 2x\sqrt{x+1} = 2x + \sqrt{x^2 + 4x + 3}\), let's start by simplifying and analyzing the terms.

First, observe that:
\[
\sqrt{x^2 + 4x + 3} = \sqrt{(x+1)(x+3)}
\]

This can be rewritten as:
\[
\sqrt{(x+1)(x+3)} = \sqrt{(x+1)} \cdot \sqrt{(x+3)}
\]

So the original equation becomes:
\[
\sqrt{x+3} + 2x\sqrt{x+1} = 2x + \sqrt{(x+1)(x+3)}
\]

Now, let's test some potential values for \(x\) to see if they satisfy the equation.

### Testing \(x = 0\):
\[
\sqrt{0+3} + 2 \cdot 0 \cdot \sqrt{0+1} = 2 \cdot 0 + \sqrt{0^2 + 4 \cdot 0 + 3}
\]
\[
\sqrt{3} + 0 = 0 + \sqrt{3}
\]
\[
\sqrt{3} = \sqrt{3}
\]
This is true, so \(x = 0\) is a solution.

### Testing \(x = 1\):
\[
\sqrt{1+3} + 2 \cdot 1 \cdot \sqrt{1+1} = 2 \cdot 1 + \sqrt{1^2 + 4 \cdot 1 + 3}
\]
\[
\sqrt{4} + 2 \cdot \sqrt{2} = 2 + \sqrt{8}
\]
\[
2 + 2\sqrt{2} = 2 + 2\sqrt{2}
\]
This is true, so \(x = 1\) is also a solution.

### Testing \(x = -1\):
\[
\sqrt{-1+3} + 2 \cdot (-1) \cdot \sqrt{-1+1} = 2 \cdot (-1) + \sqrt{(-1)^2 + 4 \cdot (-1) + 3}
\]
\[
\sqrt{2} + 2 \cdot (-1) \cdot \sqrt{0} = -2 + \sqrt{1 - 4 + 3}
\]
\[
\sqrt{2} + 0 = -2 + \sqrt{0}
\]
\[
\sqrt{2} = -2
\]
This is false, so \(x = -1\) is not a solution.

### Testing \(x = -3\):
\[
\sqrt{-3+3} + 2 \cdot (-3) \cdot \sqrt{-3+1} = 2 \cdot (-3) + \sqrt{(-3)^2 + 4 \cdot (-3) + 3}
\]
\[
\sqrt{0} + 2 \cdot (-3) \cdot \sqrt{-2} = -6 + \sqrt{9 - 12 + 3}
\]
\[
0 + 2 \cdot (-3) \cdot \sqrt{-2} = -6 + \sqrt{0}
\]
\[
0 + 0 = -6
\]
This is false, so \(x = -3\) is not a solution.

### Conclusion:
The solutions to the equation \(\sqrt{x+3} + 2x\sqrt{x+1} = 2x + \sqrt{x^2 + 4x + 3}\) are \(x = 0\) and \(x = 1\).