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Mathematics 15 Online
OpenStudy (anonymous):

Integral of dx/1+4x^2?

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

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

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

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

OpenStudy (anonymous):

im not good with trigonometric functions

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