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OpenStudy (anonymous):
indefinite integral of x sqrt of 2x+1
help!!
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OpenStudy (anonymous):
chain rule
OpenStudy (anonymous):
derivative of x times sqrt of 2x+1 + derivative of sqrt of 2x+1 times x
OpenStudy (anonymous):
oh oops its integral LOL
OpenStudy (anonymous):
same thing
OpenStudy (anonymous):
yea lol @11
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OpenStudy (anonymous):
you sure ?? @11calcBC
OpenStudy (anonymous):
no i mean the same rule apply to integration...lol
OpenStudy (anonymous):
im lost @11calcBC
OpenStudy (anonymous):
lol try substitution
OpenStudy (anonymous):
let u = 2x + 1
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OpenStudy (anonymous):
ok I have all of that I just don't know wat goes after :o @11calcBC
OpenStudy (anonymous):
wuts ur resulting equation?
OpenStudy (anonymous):
with u instead of x
OpenStudy (anonymous):
after that it's just applying regular integration rules and ur done
OpenStudy (anonymous):
well I have u=2x+1
du=2dx
and x=(u-1)/2
so I plug in what x= ??? @11calcBC
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OpenStudy (anonymous):
well if u=2x+1, x would equal (u -1)/2 now wouldn't it?
OpenStudy (anonymous):
then multiple sqrt u with (u -1)/2, take out the 1/2 (its constant) and integrate
OpenStudy (anonymous):
oh! I think I get it! thanks! @11calcBC
OpenStudy (anonymous):
no problem mate good luck!
OpenStudy (anonymous):
thanks!! @11calcBC
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