Integral word problem check. Is this right?

Question

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width. Then find the area of the region.

Here are the constraints:

$$ y =\frac{1}{x}, y=\frac{1}{x^2}, x=2$$

The integral setup I have is:

$$\int_1^2 \frac{1}{x} - \frac{1}{x^2}$$ $$\int_1^2 x^{-1} - x^{-2}$$ What is the antiderivative of $\frac{1}{x}$?

$$ln 2 - ln 1 - \frac{1}{2}$$

Answer 1

$$\text{Required area} = \int_1^2 (x^{-1} - x^{-2}) \,\mathrm{d}x = [\ln x + x^{-1}]_1^2 = (\ln2 - \ln 1) + (2^{-1}-1^{-1}) \\ = \ln2 - \frac12$$

Answer 2

You are correct in your ways, the anti derivative of $\frac1x $ is $\ln(x)$

so the value becomes ; $I= \displaystyle\int_1^2\frac1x-\frac1{x^2}\,dx = \ln|x| +\frac1x\bigg|_1^2 = \ln(2)+\frac12-\ln1-1= \ln(2)-\frac12$