Here is the graph of our function:

First, we can consider the symmetry of the problem, and reduce our integral twice the area below the function between x = 0 and x = 1:

$$A = 2 \, \lim_{t \rightarrow 0^+} \, \int_t^1 \, \frac{1}{x^2} \, dx$$

This is a fairly simple integral to do (express the fraction as x^-2 and use the power rule). The result is

$$= 2 \, \lim_{t \rightarrow 0^+} \, \left[ \frac{-1}{x} \right]_t^1 = 2 \left( \frac{-1}{1} \right) - 2 \lim_{t \rightarrow 0^+} \frac{-1}{t}$$

Now the limit on the right is infinite, so this integral diverges, and we can't calculate a finite area.

Here is the graph of our function:

Like $f(x) = ln(x),$ this function approaches -∞ as x approaches zero from above. Our integral is

$$A = 2 \, \lim_{t \rightarrow 0^+} \, \int_t^1 \, \frac{ln(x)}{\sqrt{x}} \, dx$$

To tackle the integral, let $u = \sqrt{x},$ then $u^2 = x,$ and $du = \frac{1}{2}x^{-1/2},$ so we have (forgetting the limits for now):

$$= 2 \, \int \, ln(u^2) \, du = 4 \, \int \, ln(u) \, du.$$

Now the antiderivative of ln(x) is $x ln(x) + x + C,$ so we have

$$ \begin{align} & 4 u \, ln(u) + u \; \rightarrow \\[5pt] & \lim_{t \rightarrow 0^+} \, 4 \left[ \sqrt{x} \, ln(\sqrt{x}) - \sqrt{x} \right]_t^1 \\[5pt] &= -1 - \lim_{t \rightarrow 0^+} \, \left[ \sqrt{t} \, ln(\sqrt{t}) - \sqrt{t} \right] \end{align}$$

Now we have to be careful here because the limit

$$\lim_{t \rightarrow 0^+} t \, ln(\sqrt{t})$$

is an indeterminate form, and we have to evaluate it using L'Hopital's rule. To do that we employ a trick to make the limit an ∞/∞ limit:

$$ \begin{align} &\lim_{t \rightarrow 0^+} \, \frac{ln(\sqrt{x})}{\frac{1}{x}} \\[5pt] &= \lim_{t \rightarrow 0^+} \, \frac{{t^{1/2} \left(\frac{1}{2}\right)t^{-1/2}}}{-t^{-2}} \\[5pt] &= \lim_{t \rightarrow 0^+} \, \frac{\frac{1}{2t}}{\frac{-1}{t^2}} \\[5pt] &= lim_{t \rightarrow 0^+} \, \frac{-t}{2} = 0 \end{align}$$

In the second step, we took the derivatives of the numerator and denominator, accoroding to L'Hopital's rule. Now if we remember our -1 from earlier, we see that the area is -1 units². This is a good problem to study.

Here is the graph of our function:

This is an improper integral simply because of the infinite upper bound. The integral is

$$A = \lim_{t \rightarrow \infty} \, \int_0^t \, \frac{e^x}{e^{2x} + 1} \, dx$$

This integral can be done by substitution. Let $u = e^x,$ then $du = e^x \, dx$ and $u^2 = e^{2x}.$ We can swap limits and stick with u: when $x = 0, \; u = 1,$ and when $x \rightarrow \infty, \; u \rightarrow \infty.$ Our integral is then

$$A = \lim_{t \rightarrow \infty} \, \int_0^t \, \frac{1}{u^2 + 1} \, du$$

Now recall that $\frac{d}{dx} tan^{-1}(x) = \frac{1}{x^2 + 1},$ so we have

$$ \begin{align} A &= \lim_{t \rightarrow \infty} \, tan^{-1}(u) \\[5pt] &= \lim_{t \rightarrow \infty} \, tan^{-1}(t) - tan^{-1}(1) \\[5pt] &= \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \; units^2 \end{align}$$

The area under this curve from x=0 to x=∞ is finite.

Here is the graph of our function:

This is an improper integral because of the discontinuity at x = 1. Notice that if we just did the integral in a straightforward way, ignoring the vertical asymptote, we could do it by substitution: let $u = x - 1,$ then $du = dx$ and $x = u + 1,$ so we have

$$ \begin{align} A &= \int_{-1}^4\, \frac{u + 1}{u} \, du \\[5pt] &= \int_{-1}^4\, 1 + \frac{1}{u} \, du \\[5pt] &= u + ln|u| \, \bigg|_{-1}^4 \\[5pt] &= 4 + ln(4) + 1 - ln|-1| = 5 + ln(4) \end{align}$$

But this is not correct, as we shall see. The proper improper integral is

$$A = \lim_{t \rightarrow 1} \, \int_0^t \, \frac{x}{x - 1} \, dx + \lim_{t \rightarrow 1} \, \int_t^5 \, \frac{x}{x - 1} \, dx$$

Using the same substitution and swapping limits to u-limits, we have

$$ \begin{align} &\lim_{t \rightarrow 1} \int_{-1}^{t-1} \left(1 + \frac{1}{u} \right) du + \lim_{t \rightarrow 1} \int_{t-1}^{4} \left(1 + \frac{1}{u} \right) du\\[5pt] &= \lim_{t \rightarrow 1} u + ln|u|\bigg|_{-1}^{t-1} + \lim_{t \rightarrow 1} u + ln|u| \bigg|_{t-1}^4 \end{align}$$

This expression will contain two terms like

$$\lim_{t \rightarrow 1} \, ln|t - 1|,$$

so the limit is infinite, as is the area. The integral diverges.

Here is the graph of our function:

This is an improper integral because of the infinite upper limit. The improper integral is:

$$\lim_{t \rightarrow \infty} \, \int_0^t x \, e^{-3x} \, dx$$

We can do this integral by parts. Let $u = x,$ and $dv = e^{-3x} dx.$ Then $du = dx$ and $v = -\frac{1}{3} e^{-3x}.$ Then our integral becomes (forgetting the limits for now):

$$\lim_{t \rightarrow \infty} \, -\frac{x}{3} e^{-3x} + \frac{1}{3} \int e^{-3x} \, dx$$

Integrating and including the limits gives

$$= \lim_{t \rightarrow \infty} \, -\frac{x}{3} e^{-3x} - \frac{1}{9} e^{-3x} \, \bigg|_0^t.$$

Before evaluating the limits, it might be enlightening to rewrite that expression slightly, like this:

$$= \lim_{t \rightarrow \infty} \, -\frac{x}{3 e^{3x}} - \frac{1}{9 e^{3x}} \, \bigg|_0^t.$$

Now we can evaluate:

$$\lim_{t \rightarrow \infty} \, \frac{-t}{3 e^{3t}} - \frac{1}{9 e^{3t}} + 0 + \frac{1}{9}$$

The first limit can be evaluated using L'Hopital's rule. It reduces to zero. So our overal result is $\frac{1}{9}.$

Here is the graph of our function:

This is an improper integral because of the infinite limits. Because of the symmetry of the function, we can just integrate from 0 to ∞ and multiply by 2. The improper integral is:

$$2 \lim_{t \rightarrow \infty} \, \int_0^t \frac{1}{1 + x^2} \, dx$$

This is an inverse tangent integral. Recall that $\frac{d}{dx} tan^{-1}(x) = \frac{1}{x^2 + 1},$ so we have

$$2 \lim_{t \rightarrow \infty} \, tan^{-1}(x) \, \bigg|_0^t$$

The inverse tangent of zero is zero, and the inverse tangent of ∞ is π/2, so our limit is

$$= 2\lim_{t \rightarrow \infty} \, tan^{-1}(t) - tan^{-1}(0) = \pi \; units^2$$