Evaluate the definite integral.

The integral to evaluate is $\int_1^p (px-1) dx$.
First, find the antiderivative of $px-1$ with respect to $x$:

Now, apply the limits of integration from 1 to $p$:

Combine the terms involving $p$:

We are given that $\int_1^p (px-1) dx = 0$.
So, we set the expression from Step 1 to 0:

Multiply the entire equation by 2 to eliminate the denominators:

We look for integer roots by testing divisors of the constant term (2), which are $\pm 1, \pm 2$.
Test $p=1$:

Since $p=1$ satisfies the equation, $(p-1)$ is a factor of the polynomial.
We can use polynomial division or synthetic division to find the other factors. Using synthetic division with root 1:

1 | 1   0   -3   2
  |     1    1  -2
  ----------------
    1   1   -2   0

The quotient is $p^2 + p - 2$.
So, the cubic equation can be factored as:

Now, factor the quadratic term $p^2 + p - 2$. We need two numbers that multiply to -2 and add to 1. These numbers are 2 and -1.

Substitute this back into the factored cubic equation:

From $(p-1)^2 = 0$, we get $p-1 = 0$, which means $p=1$.
From $(p+2) = 0$, we get $p=-2$.

Thus, the values of $p$ are 1 and $-2$.

The final answer is $p=1, p=-2$.