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Mathematics 14 Online
OpenStudy (anonymous):

Integral Help :)

OpenStudy (anonymous):

\[\int\limits_{}^{}(zdz)/(\sqrt{z-2})\]

OpenStudy (anonymous):

I set u as z-2....

OpenStudy (anonymous):

seems like a good idea

OpenStudy (anonymous):

\[\int\limits\limits_{}^{}(u+2)du * 1/\sqrt{u}\]

OpenStudy (anonymous):

So you would get something like that?

OpenStudy (anonymous):

maybe \[\int\frac{u+2}{\sqrt{u}}du\]

OpenStudy (anonymous):

rewrite in two pieces with exponents, use the backwards power rule

OpenStudy (anonymous):

Hmmm? I don't understand

OpenStudy (zale101):

\[\int\limits_{}^{}\frac{u}{\sqrt{u}}+\frac{2}{\sqrt{u}}~du\]

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