It suggests the integral may converge, but further analysis is needed. The function must approach 0 'fast enough' for convergence.

The integral likely diverges, as the area under the curve does not settle to a finite value.

Look for vertical asymptotes or holes within the integration interval.

The value of the definite integral is positive.

The value of the definite integral is negative.

If the function decreases rapidly, it is more likely to converge. If it decreases slowly or oscillates, it is more likely to diverge.

If the integral converges, it will converge to a positive value.

If the integral converges, it will converge to a negative value.

By visually estimating the area under the curve and considering the behavior as x approaches infinity or a point of discontinuity.

The integral must be split at $x=c$ and evaluated as two separate improper integrals using limits.

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

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

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

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

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

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

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

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

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

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

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 $\frac{1}{2}$.