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OpenStudy (anonymous):
integrate x^2/ sqrt(4+x^2)
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hartnn (hartnn):
did you try
u= 4+x^2
?
OpenStudy (anonymous):
no, let me do that
hartnn (hartnn):
cool, let me know if you get stuck on any step :)
hartnn (hartnn):
another way i could think of is to write
x^2 as
x^2+4-4
and then splitting the fraction
sqrt (x^2+4)-4/sqrt(x^2+4)
OpenStudy (anonymous):
du=2xdx and im not sure what to do with the x
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hartnn (hartnn):
x= \sqrt (u-4)
now find dx
OpenStudy (anonymous):
my new integral looks like \[\int\limits (u-4)/ 2(\sqrt{u \sqrt{u-4}})\]
OpenStudy (xapproachesinfinity):
xdx=du
you don't need to do anything
you x^2 on top
make it x.xdx
OpenStudy (xapproachesinfinity):
However, this is little different from what @hartnn suggested
OpenStudy (anonymous):
so what would u be?
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OpenStudy (anonymous):
Some other suggestions for substituting: try \(x=2\tan u\) or \(x=\sqrt t\).
OpenStudy (xapproachesinfinity):
oh yeah i like that sub @SithsAndGiggles suggested just now
makes thing a lot simpler
try it^_^
OpenStudy (xapproachesinfinity):
i meant the x=2tanu
OpenStudy (xapproachesinfinity):
this requires some trig identities to do^_^
OpenStudy (anonymous):
how do i make x=2tanu?
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OpenStudy (xapproachesinfinity):
for u as you started first
u=4+x^2
du=2xdx =====> 1/2du=xdx if you want to continue this way
OpenStudy (xapproachesinfinity):
set x=2tanu
then x^2=4tan^2u
OpenStudy (xapproachesinfinity):
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