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anonymous
 5 years ago
hi i need help finding the integral of this of 3/x^2+4
anonymous
 5 years ago
hi i need help finding the integral of this of 3/x^2+4

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do you mean\[\int\limits_{}{}\frac{3}{x^2+4}dx\]?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Make a substitution of x=2tan(x) and go from there.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i totally for got can u please walk me through it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let x=2tan(theta). Then\[\int\limits{}{}\frac{3}{x^2+4}dx=3\int\limits{}{}\frac{d(2\tan \theta)}{4\tan^2 \theta +4}=\frac{3}{4}\int\limits{}{}\frac{2\sec ^2 \theta d \theta}{1+\tan^2 \theta}\]\[=\frac{6}{4}\int\limits{}{}\frac{\sec ^2 \theta d \theta}{\sec^2 \theta}=\frac{3}{2}\int\limits{}{} d \theta= \frac{3}{2} \theta+c\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Since\[x = 2\tan \theta \rightarrow \theta = \tan^{1}\frac{x}{2}\]so\[\int\limits_{}{}\frac{3}{x^2+4}dx= \frac{3}{2}\tan^{1}\frac{x}{2}+c\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohh the limits of integration are infinity to 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, you just can use the result ^^. The constant will cancel. In improper integrals, you should set the problem up as\[\lim_{c \rightarrow \infty}\int\limits_{1}^{c}\frac{3}{x^2+4}dx=\lim_{c \rightarrow \infty}\frac{3}{2}\tan^{1}\frac{x}{2}_1^c\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[=\lim_{c \rightarrow \infty}\left[ \frac{3}{2}\tan^{1}\frac{c}{2}\frac{3}{2}\tan^{1}\frac{1}{2} \right]\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[=\lim_{c \rightarrow \infty}\frac{3}{2}\tan ^{1}\frac{c}{2}\frac{3}{2}\tan^{1}\frac{1}{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[=\frac{3}{2}.\frac{\pi}{2}\frac{3}{2}\tan^{1}\frac{1}{2}=3\pi  \frac{3}{2}\tan^{1} \frac{1}{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey thank u so much for ur help . can i ask u another quick question

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Wait, the calculation of the limits is wrong.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey srry i just checked its 0 to infinity

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the anser is 3pi/4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, the limit of arctan at infinity is pi/2, and arctan of 0 is 0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The answer will be 3pi.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The working for the integral is correct. Change your limits as needed.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0kk great ty can i ask u something very quick about another problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why does this diverge \[\sum_{2}^{\infty} \] n/n+1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Since the limit of each term of your sequence does not go to zero, by the nth term test, the series will diverge. If \[\lim_{n \rightarrow \infty}a_n \ne 0\]then\[\sum a_n\] will diverge.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0DON'T CONFUSE THIS WITH THE FOLLOWING: If \[\lim_{n \rightarrow \infty} a_n = 0 \]then\[\sum a_n\]converges.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im still confused bc if u plug in infinity u get infinity/infinity

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0^^ is not always true

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{n}{n+1}=\frac{1}{1+1/n}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now send n to infinity.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did u get 1/1+1/n

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{n \rightarrow \infty}\frac{n}{n+1} =\lim_{n \rightarrow \infty}\frac{1}{1+1/n}=\frac{1}{1+0}=1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Divide the numerator AND denominator by n (at the same time). A common practice is to divide the numerator AND denominator with the highest powered term in a polynomial quotient.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then why is the answer divergent if u got 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can also think of it like this:\[\frac{n}{n+1}=\frac{n}{n(1+1/n)}=\frac{1}{1+1/n}\]since the n's cancel.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The theorem says, IF you DON'T get 0 when you take the limit of the sequence that makes up the series, then it WON'T converge. 1 is not 0. What it boils down to is a series whose terms aren't dying quick enough for everything to settle down and converge to a limit.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh yah i totally forgot

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Don't ever think, though, you can take the limit of a_n and get 0 and your series converges. That's a trap. It ONLY says that if it DOESN'T go to ZERO, then it DEFINITELY WON'T converge.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the answer is also 1 bc if the n of the top and bottom are to the same power then u can take into account the coefficients

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm not too sure what you mean.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lim as n goes to infinity n/n the answer is 1/1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think you need to hone in your algebra skills before moving too much further with sequences and series, since this part of mathematics is all about algebraic manipulation/tricks.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, since n/n = 1 which is constant.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But I think you're talking about something called 'asymptotic equivalence'. In the limit, n+1 is approximately equal to n and so it can be expected that, for large n, n/(n+1) is approximately equal to n/n which is 1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yah something like that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The limit as n goes to infinity of 1 is always 1 since it's independent of n.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Try this site: http://tutorial.math.lamar.edu/Classes/CalcII/SeriesIntro.aspx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The attachment I sent is on asymptotic equivalence. The method is very helpful when trying to find limits.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0k ty okay im solving another problem and \[\sum_{1}^{\infty}\] 1/2^n+1/n and i don"t know why this is divergent bc when i plug in infi i get 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It goes back to what I was saying before  going to 0 is no guarantee of convergence. Only NOT GOING TO ZERO is a guarantee of DIVERGENCE.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0k so i know 1/n does bc the power of n is greater than zero but how do i figure out the integral of 2^n

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0isnt the integral of 2^n ln2^n

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0srry i meant ln2(2^n)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You don't need to use integral test. Use limit comparison. Use the harmonic series as your test series to find that\[\lim_{n \rightarrow \infty}\frac{1/n}{1/2^n+1/n}=\lim_{n \rightarrow \infty}\frac{2^n}{2^n+n}=\lim_{n \rightarrow \infty}\frac{2^n}{2^n(1+n/2^n)}\]\[=\lim_{n \rightarrow \infty}\frac{1}{1+n/2^n}=\lim_{n \rightarrow \infty}\frac{1}{1+0}=1\]since this is a nonzero limit, and your test series is a known divergent series, the limit comparison test says that your own series must diverge too.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0alright thank u i got it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Get your algebra sorted, and use online resources!
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