Mathematics
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OpenStudy (anonymous):
Integral of dx/1+4x^2?
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OpenStudy (anonymous):
\[\int\limits \frac{ dx }{ 1+4x ^{2} }\]
OpenStudy (anonymous):
@e.mccormick @jim_thompson5910
jimthompson5910 (jim_thompson5910):
check out this table of integrals
http://integral-table.com/
look under "Integrals of Rational Functions"
jimthompson5910 (jim_thompson5910):
You'll have to do a bit of substitution though. Let u = 2x, so u^2 = 4x^2
OpenStudy (anonymous):
oh ok
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OpenStudy (anonymous):
so u=2x
jimthompson5910 (jim_thompson5910):
don't forget to calculate du/dx and somehow figure out a way to replace dx as well
OpenStudy (anonymous):
\[\frac{ 1 }{ 2 }\int\limits \frac{ 1 }{ 1+u ^{2} }du \]
jimthompson5910 (jim_thompson5910):
looks good
jimthompson5910 (jim_thompson5910):
now you use that table
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OpenStudy (anonymous):
sorry my internet just crashed
jimthompson5910 (jim_thompson5910):
thats fine
OpenStudy (anonymous):
\[\frac{ 1 }{ 2 }\tan ^{-1}u\]
jimthompson5910 (jim_thompson5910):
recall that you let u = 2x, so replace u with 2x and don't forget the +C
jimthompson5910 (jim_thompson5910):
also, say arctan(u) and NOT arctan u
the parenthesis help show you where the function ends
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OpenStudy (anonymous):
\[\frac{ 1 }{ 2 }\arctan(2x) +C\]
jimthompson5910 (jim_thompson5910):
good
jimthompson5910 (jim_thompson5910):
you nailed it
OpenStudy (anonymous):
ok, thanks
jimthompson5910 (jim_thompson5910):
np
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OpenStudy (anonymous):
im not good with trigonometric functions