Type 1 integrals with part c
Evaluate, $\int_{-\infty}^{\infty}\frac{8}{1+x^2}dx$

Determining convergence and divergence with type 2 integrals
Evaluate,

$\int_{0}^{\pi}\frac{\sin^2x}{\cos x}dx$

$\int_{0}^{e}\ln x$$dx$

$\int_{2}^{3}\frac{1}{\sqrt{3-x}}dx$

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First, we will learn about Type 1 improper integrals. These types of improper integrals have bounds which have positive or negative infinity. Then we will look at Type 2 improper integrals. These improper integrals happen when the function is undefined at a specific place or area within the region of integration. For these integrals, we will have to use limits. If the limit exists and is finite, then the integral can be solved. Otherwise, the integral will be unsolvable.

Note:
There are two types of improper integrals:
1) Type 1
a) $\int_{a}^{\infty}f(x) dx$$=$$\lim$_{t →$\infty$}$\int_{a}^{t}f(x)dx$

b) $\int_{-\infty}^{b}f(x)dx=$$\lim$_{t →$-\infty$}$\int_{t}^{b}f(x)dx$

c) $\int_{-\infty}^{\infty}f(x)dx=\int_{-\infty}^{a}f(x)dx+\int_{a}^{\infty}f(x)dx$

2) Type 2
a) If $f$ is continuous on $[a,b)$ and discontinuous at $b$, then:
$\int_{a}^{b} f(x)dx=$$\lim$_{t →$b^-$}$\int_{a}^{t}f(x)dx$

b) If $f$ is continuous on $(a,b]$ and discontinuous at $a$, then:
$\int_{a}^{b} f(x)dx=$$\lim$_{t →$a^+$}$\int_{t}^{b}f(x)dx$

c) If $f$ has a discontinuity at $c$, where $a<c<b$, then: $\int_{a}^{b} f(x)dx=\int_{a}^{c} f(x)dx+\int_{c}^{b} f(x)dx$

If the limits exist and is finite, then it is convergent. Otherwise, it is divergent.