### Instruction: Find all values of $x$ that satisfy
\[5x - 1 < (x + 1)^2 < 7x - 3.\].### Response: This problem involves inequalities, so I need to be careful with the signs and the intervals.
I can try to simplify each inequality separately by expanding and rearranging the terms.
For the left inequality, I get
\[(x + 1)^2 > 5x - 1 \iff x^2 + 2x - 5x + 1 + 1 > 0 \iff x^2 - 3x + 2 > 0.\]
To find the values of $x$ that make this quadratic expression positive, I can factor it and use a sign diagram.
The expression factors as $(x - 2)(x - 1)$, so the zeros are $x = 2$ and $x = 1$.
A sign diagram shows that the expression is positive when $x < 1$ or $x > 2$.
So the left inequality is true for $x < 1$ or $x > 2$.
For the right inequality, I get
\[(x + 1)^2 < 7x - 3 \iff x^2 + 2x - 7x + 1 + 3 < 0 \iff x^2 - 5x + 4 < 0.\]
Again, I can factor this expression and use a sign diagram.
The expression factors as $(x - 4)(x - 1)$, so the zeros are $x = 4$ and $x = 1$.
A sign diagram shows that the expression is negative when $1 < x < 4$.
So the right inequality is true for $1 < x < 4$.
Now I have to combine the two inequalities and find the common values of $x$ that satisfy both.
This means I have to find the intersection of the intervals $x < 1$ or $x > 2$ and $1 < x < 4$.
I can do this by drawing a number line and shading the regions that correspond to each inequality.
The intersection is the region that is shaded by both inequalities, which is $2 < x < 4$.
So the final answer is $2 < x < 4$.