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ksaimouli
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http://online.math.uh.edu/apcalculus/exams/images/AP_AB_version1__37.gif
 one year ago
 one year ago
ksaimouli Group Title
http://online.math.uh.edu/apcalculus/exams/images/AP_AB_version1__37.gif
 one year ago
 one year ago

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ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
@mathmate
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
i did this with u substituion it did not work
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
i am getting something like ln2/2
 one year ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
It looks to me more an atan(2x)
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
i took u=1+4t^2
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
du/8t=dt
 one year ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
The standard integral is \( \int \frac{dx}{1+x^2} = atan(x) + C \)
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
dw:1358025284718:dw
 one year ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
So you'd be integrating \( \int \frac{du}{8u(1+4t^2)} \)
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
use the equation button
 one year ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
The result is not ln(u)
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
ok i should remember that formula right?
 one year ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
I suggest you use y=2t
 one year ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
Yes, it's on the first page of standard integrals. Check it out.
 one year ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
\( \large \int_0^{1/2} \frac{dt}{1+4t^2}= \int_0^{1} \frac{2dy}{1+y^2} = [atan(y)]_0^{1}= 2\pi/4=\pi/2 \)
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
@mathmate i found the formula to be \[\frac{ 1 }{ a }\tan^{1} \frac{ x }{ a }+c\]
 one year ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
Right, so a=1/2.
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
what is x then
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{}^{}\frac{ 1 }{ a^2+x^2}=\frac{ 1 }{ a }\tan^{1} \frac{ x }{ a}+c\]
 one year ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
x is your t. you can replace the identity with \( \large \int \frac{dx}{a^2+x^2} = \int \frac{dt}{a^2(1+(t/a)^2)} \)
 one year ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
and a=1/2
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
can i do \[\frac{ 1 }{ 1 }\tan^{1} \frac{ 2t }{ 1}\]
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
and t=1/2
 one year ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
t is a variable, and you cannot equate it to a constant. Use the formula you found, change t for x, and 1/2 for a, then the lefthand side is exactly your question. All you need to do after that is to evaluate the righthand side.
 one year ago
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