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

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

Mathematics
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Do you mean\[\int\limits_{}{}\frac{3}{x^2+4}dx\]?
yah
-3/x+4x +Constant

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Make a substitution of x=2tan(x) and go from there.
i totally for got can u please walk me through it
ok
Let x=2tan(theta). Then\[\int\limits{}{}\frac{3}{x^2+4}dx=3\int\limits{}{}\frac{d(2\tan \theta)}{4\tan^2 \theta +4}=\frac{3}{4}\int\limits{}{}\frac{2\sec ^2 \theta d \theta}{1+\tan^2 \theta}\]\[=\frac{6}{4}\int\limits{}{}\frac{\sec ^2 \theta d \theta}{\sec^2 \theta}=\frac{3}{2}\int\limits{}{} d \theta= \frac{3}{2} \theta+c\]
Since\[x = 2\tan \theta \rightarrow \theta = \tan^{-1}\frac{x}{2}\]so\[\int\limits_{}{}\frac{3}{x^2+4}dx= \frac{3}{2}\tan^{-1}\frac{x}{2}+c\]
ohh the limits of integration are infinity to 1
Well, you just can use the result ^^. The constant will cancel. In improper integrals, you should set the problem up as\[\lim_{c \rightarrow \infty}\int\limits_{1}^{c}\frac{3}{x^2+4}dx=\lim_{c \rightarrow \infty}\frac{3}{2}\tan^{-1}\frac{x}{2}|_1^c\]
\[=\lim_{c \rightarrow \infty}\left[ \frac{3}{2}\tan^{-1}\frac{c}{2}-\frac{3}{2}\tan^{-1}\frac{1}{2} \right]\]
\[=\lim_{c \rightarrow \infty}\frac{3}{2}\tan ^{-1}\frac{c}{2}-\frac{3}{2}\tan^{-1}\frac{1}{2}\]
\[=\frac{3}{2}.\frac{\pi}{2}-\frac{3}{2}\tan^{-1}\frac{1}{2}=3\pi - \frac{3}{2}\tan^{-1} \frac{1}{2}\]
hey thank u so much for ur help . can i ask u another quick question
Wait, the calculation of the limits is wrong.
hey srry i just checked its 0 to infinity
so the anser is 3pi/4
Well, the limit of arctan at infinity is pi/2, and arctan of 0 is 0.
The answer will be 3pi.
k ty
The working for the integral is correct. Change your limits as needed.
kk great ty can i ask u something very quick about another problem
if it's quick!
why does this diverge \[\sum_{2}^{\infty} \] n/n+1
Since 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.
DON'T CONFUSE THIS WITH THE FOLLOWING: If \[\lim_{n \rightarrow \infty} a_n = 0 \]then\[\sum a_n\]converges.
im still confused bc if u plug in infinity u get infinity/infinity
^^ is not always true
\[\frac{n}{n+1}=\frac{1}{1+1/n}\]
Now send n to infinity.
how did u get 1/1+1/n
\[\lim_{n \rightarrow \infty}\frac{n}{n+1} =\lim_{n \rightarrow \infty}\frac{1}{1+1/n}=\frac{1}{1+0}=1\]
Divide 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.
then why is the answer divergent if u got 1
You 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.
The 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.
oh yah i totally forgot
Don'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.
the 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
I'm not too sure what you mean.
lim as n goes to infinity n/n the answer is 1/1
I 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.
Yes, since n/n = 1 which is constant.
But 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.
yah something like that
The limit as n goes to infinity of 1 is always 1 since it's independent of n.
Try this site: http://tutorial.math.lamar.edu/Classes/CalcII/SeriesIntro.aspx
1 Attachment
The attachment I sent is on asymptotic equivalence. The method is very helpful when trying to find limits.
k 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
It 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.
k 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
isnt the integral of 2^n ln2^n
srry i meant ln2(2^n)
You 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 non-zero limit, and your test series is a known divergent series, the limit comparison test says that your own series must diverge too.
alright thank u i got it
Good :p
Get your algebra sorted, and use online resources!
k

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