1. Express as a limit: $\lim_{b \to \infty} \int_0^b e^{-x} dx$. 2) Evaluate the integral: $\lim_{b \to \infty} [-e^{-x}]_0^b = \lim_{b \to \infty} (-e^{-b} + e^0)$. 3) Evaluate the limit: $\lim_{b \to \infty} (-e^{-b} + 1) = 1$. 4) The integral converges to 1.

1. Express as a limit: $\lim_{b \to \infty} \int_1^b \frac{1}{x^2} dx$. 2) Evaluate the integral: $\lim_{b \to \infty} [-\frac{1}{x}]_1^b = \lim_{b \to \infty} (-\frac{1}{b} + 1)$. 3) Evaluate the limit: $\lim_{b \to \infty} (-\frac{1}{b} + 1) = 1$. 4) The integral converges to 1.

1. Express as a limit: $\lim_{a \to 0^+} \int_a^1 \frac{1}{\sqrt{x}} dx$. 2) Evaluate the integral: $\lim_{a \to 0^+} [2\sqrt{x}]_a^1 = \lim_{a \to 0^+} (2 - 2\sqrt{a})$. 3) Evaluate the limit: $\lim_{a \to 0^+} (2 - 2\sqrt{a}) = 2$. 4) The integral converges to 2.

1. Express as a limit: $\lim_{b \to \infty} \int_1^b \frac{1}{x} dx$. 2) Evaluate the integral: $\lim_{b \to \infty} [\ln|x|]_1^b = \lim_{b \to \infty} (\ln(b) - \ln(1))$. 3) Evaluate the limit: $\lim_{b \to \infty} (\ln(b) - 0) = \infty$. 4) The integral diverges.

1. Express as a limit: $\lim_{a \to -\infty} \int_a^0 xe^x dx$. 2) Integrate by parts: $u=x, dv=e^x dx$, so $du=dx, v=e^x$. Then $\int xe^x dx = xe^x - \int e^x dx = xe^x - e^x + C$. 3) Evaluate the integral: $\lim_{a \to -\infty} [xe^x - e^x]_a^0 = \lim_{a \to -\infty} [(0 - e^0) - (ae^a - e^a)] = \lim_{a \to -\infty} [-1 - ae^a + e^a]$. 4) Evaluate the limit: $\lim_{a \to -\infty} [-1 - ae^a + e^a] = -1 - 0 + 0 = -1$. 5) The integral converges to -1.

1. Split the integral: $\int_{-\infty}^{\infty} \frac{1}{1+x^2} dx = \int_{-\infty}^{0} \frac{1}{1+x^2} dx + \int_{0}^{\infty} \frac{1}{1+x^2} dx$. 2) Express as limits: $\lim_{a \to -\infty} \int_a^0 \frac{1}{1+x^2} dx + \lim_{b \to \infty} \int_0^b \frac{1}{1+x^2} dx$. 3) Evaluate the integral: $\lim_{a \to -\infty} [\arctan(x)]_a^0 + \lim_{b \to \infty} [\arctan(x)]_0^b = \lim_{a \to -\infty} [\arctan(0) - \arctan(a)] + \lim_{b \to \infty} [\arctan(b) - \arctan(0)]$. 4) Evaluate the limits: $[0 - (-\frac{\pi}{2})] + [\frac{\pi}{2} - 0] = \frac{\pi}{2} + \frac{\pi}{2} = \pi$. 5) The integral converges to $\pi$.

1. Express as a limit: $\lim_{b \to \infty} \int_2^b \frac{1}{x(x-1)} dx$. 2) Partial fraction decomposition: $\frac{1}{x(x-1)} = \frac{A}{x} + \frac{B}{x-1}$. Solving gives $A = -1$ and $B = 1$. So, $\frac{1}{x(x-1)} = \frac{-1}{x} + \frac{1}{x-1}$. 3) Evaluate the integral: $\lim_{b \to \infty} \int_2^b (\frac{-1}{x} + \frac{1}{x-1}) dx = \lim_{b \to \infty} [-\ln|x| + \ln|x-1|]_2^b = \lim_{b \to \infty} [\ln|\frac{x-1}{x}|]_2^b = \lim_{b \to \infty} [\ln|\frac{b-1}{b}| - \ln|\frac{2-1}{2}|]$. 4) Evaluate the limit: $\lim_{b \to \infty} [\ln|\frac{b-1}{b}| - \ln(\frac{1}{2})] = \ln(1) - \ln(\frac{1}{2}) = 0 - (-\ln(2)) = \ln(2)$. 5) The integral converges to $\ln(2)$.

1. Express as a limit: $\lim_{b \to \infty} \int_0^b cos(x) dx$. 2) Evaluate the integral: $\lim_{b \to \infty} [sin(x)]_0^b = \lim_{b \to \infty} (sin(b) - sin(0))$. 3) Evaluate the limit: $\lim_{b \to \infty} (sin(b) - 0)$. Since $sin(b)$ oscillates between -1 and 1 as b approaches infinity, the limit does not exist. 4) The integral diverges.

1. Split the integral at the discontinuity: $\int_0^{3} \frac{1}{x-2} dx = \int_0^{2} \frac{1}{x-2} dx + \int_2^{3} \frac{1}{x-2} dx$. 2) Express as limits: $\lim_{b \to 2^-} \int_0^b \frac{1}{x-2} dx + \lim_{a \to 2^+} \int_a^3 \frac{1}{x-2} dx$. 3) Evaluate the integral: $\lim_{b \to 2^-} [\ln|x-2|]_0^b + \lim_{a \to 2^+} [\ln|x-2|]_a^3 = \lim_{b \to 2^-} [\ln|b-2| - \ln|-2|] + \lim_{a \to 2^+} [\ln|3-2| - \ln|a-2|] = \lim_{b \to 2^-} [\ln|b-2| - \ln(2)] + \lim_{a \to 2^+} [\ln(1) - \ln|a-2|]$. 4) Evaluate the limits: Since $\lim_{b \to 2^-} \ln|b-2| = -\infty$ and $\lim_{a \to 2^+} \ln|a-2| = -\infty$, both integrals diverge. 5) The integral diverges.

1. Express as a limit: $\lim_{b \to \infty} \int_0^b \frac{x}{(1+x^2)^2} dx$. 2) Use u-substitution: Let $u = 1+x^2$, then $du = 2x dx$, so $x dx = \frac{1}{2} du$. The integral becomes $\frac{1}{2} \int \frac{1}{u^2} du = \frac{1}{2} \int u^{-2} du = \frac{1}{2} [-\frac{1}{u}] + C = -\frac{1}{2(1+x^2)} + C$. 3) Evaluate the integral: $\lim_{b \to \infty} [-\frac{1}{2(1+x^2)}]_0^b = \lim_{b \to \infty} [-\frac{1}{2(1+b^2)} - (-\frac{1}{2(1+0^2)})] = \lim_{b \to \infty} [-\frac{1}{2(1+b^2)} + \frac{1}{2}]$. 4) Evaluate the limit: $\lim_{b \to \infty} [-\frac{1}{2(1+b^2)} + \frac{1}{2}] = 0 + \frac{1}{2} = \frac{1}{2}$. 5) The integral converges to $\frac{1}{2}$.

$\int_a^\infty f(x) , dx = \lim_{b \to \infty} \int_a^b f(x) , dx$

$\int_{-\infty}^b f(x) , dx = \lim_{a \to -\infty} \int_a^b f(x) , dx$

$\int_{-\infty}^\infty f(x) , dx = \lim_{a \to -\infty} \int_a^c f(x) , dx + \lim_{b \to \infty} \int_c^b f(x) , dx$

$\int \frac{1}{x} dx = \ln|x| + C$

$\int \frac{1}{a^2+x^2} dx = \frac{1}{a} \arctan(\frac{x}{a}) + C$

$\frac{P(x)}{Q(x)} = \frac{A}{x-a} + \frac{B}{x-b} + ...$

$\int f(g(x))g'(x)dx = \int f(u)du$ where $u=g(x)$

$\int e^x dx = e^x + C$

$\int x^n dx = \frac{x^{n+1}}{n+1} + C$, where $n \neq -1$

$\int sin(x) dx = -cos(x) + C$

If $F(x) = \int_a^x f(t) dt$, then $F'(x) = f(x)$.

If $g(x) \leq f(x) \leq h(x)$ and $\int_a^\infty g(x) dx$ and $\int_a^\infty h(x) dx$ both converge to the same limit, then $\int_a^\infty f(x) dx$ also converges to that limit.

If $0 \leq f(x) \leq g(x)$ for all $x \geq a$, then if $\int_a^\infty g(x) dx$ converges, so does $\int_a^\infty f(x) dx$, and if $\int_a^\infty f(x) dx$ diverges, so does $\int_a^\infty g(x) dx$.

$\int_a^b F'(x) dx = F(b) - F(a)$

If $\lim_{x \to \infty} \frac{f(x)}{g(x)} = c$, where $0 < c < \infty$, then $\int_a^\infty f(x) dx$ and $\int_a^\infty g(x) dx$ either both converge or both diverge.

$\int u dv = uv - \int v du$

$\int f(g(x))g'(x)dx = \int f(u)du$ where $u=g(x)$

Decompose a rational function into simpler fractions that are easier to integrate.

$\int \frac{1}{x} dx = \ln|x| + C$

$\int e^x dx = e^x + C$