I have to take the integral of this: x/(sqrt(9- x^2)).
The first step is to rewrite the integral as (-1/2)the integral of (9 - x^2)^(-1/2) (______x)dx. What goes in the space (_____x)?

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OpenStudy (anonymous):

\[\int\frac{x}{\sqrt{9-x^2}}dx\]

OpenStudy (anonymous):

yep that's the integral

myininaya (myininaya):

let u=9-x^2
du=-2xdx
-1/2 du=x dx

OpenStudy (anonymous):

yep, I have that

OpenStudy (anonymous):

straight forward u-sub. put
\[u=1-x^2\]
\[du=-2xdx\]

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OpenStudy (anonymous):

if you are answering a web question step by step it looks like you should fill in the blank with -2

OpenStudy (anonymous):

oh, just put the u-sub in the space? with (-1/2) on the ouside?

OpenStudy (anonymous):

oh, okay thanks

myininaya (myininaya):

\[-1/2*\int\limits_{}^{}\frac{du}{\sqrt{u}}\]

OpenStudy (anonymous):

your question was "what goes in the blank?"
it is -2

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OpenStudy (anonymous):

but why does that work?

OpenStudy (anonymous):

the answer works

OpenStudy (anonymous):

backwards chain rule

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

\[\frac{d}{dx} f(u(x))=f'(u(x))u'(x)\]

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