Rewrite the integrand using exponent rules.

The integral is given as $\int_{x=1}^{x=2} \left(\frac{1}{x} - x^2\right) dx$.
We can rewrite $\frac{1}{x}$ as $x^{-1}$.

$$\int_{1}^{2} \left(x^{-1} - x^2\right) dx$$

The integral of $x^{-1}$ is $\ln|x|$.
The integral of $x^n$ is $\frac{x^{n+1}}{n+1}$. So, the integral of $-x^2$ is $-\frac{x^{2+1}}{2+1} = -\frac{x^3}{3}$.

$$\left[\ln|x| - \frac{x^3}{3}\right]_{1}^{2}$$

Substitute the upper limit ($x=2$) and subtract the result of substituting the lower limit ($x=1$).

$$\left(\ln|2| - \frac{2^3}{3}\right) - \left(\ln|1| - \frac{1^3}{3}\right)$$ $$\left(\ln(2) - \frac{8}{3}\right) - \left(\ln(1) - \frac{1}{3}\right)$$

Since $\ln(1) = 0$:

$$\left(\ln(2) - \frac{8}{3}\right) - \left(0 - \frac{1}{3}\right)$$ $$\ln(2) - \frac{8}{3} + \frac{1}{3}$$ $$\ln(2) - \frac{7}{3}$$