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Do you mean\[\int\limits_{}{}\frac{3}{x^2+4}dx\]?

yah

-3/x+4x +Constant

Make a substitution of x=2tan(x) and go from there.

i totally for got can u please walk me through it

ok

ohh the limits of integration are infinity to 1

\[=\lim_{c \rightarrow \infty}\frac{3}{2}\tan ^{-1}\frac{c}{2}-\frac{3}{2}\tan^{-1}\frac{1}{2}\]

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

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

then why is the answer divergent if u got 1

oh yah i totally forgot

I'm not too sure what you mean.

lim as n goes to infinity n/n the answer is 1/1

Yes, since n/n = 1 which is constant.

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

isnt the integral of 2^n ln2^n

srry i meant ln2(2^n)

alright thank u i got it

Good :p

Get your algebra sorted, and use online resources!