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OpenStudy (anonymous):

what's the indefinite integral of y=(e^(x/2)) / (sqrtx) ?

OpenStudy (anonymous):

i meant square root of x in the sqrtx part

myininaya (myininaya):

this is not an elementary integral what class is this?

OpenStudy (anonymous):

i'm in precalculus honors

myininaya (myininaya):

precalculus and integrals? lol

OpenStudy (anonymous):

idk my teacher wants us to learn more caluclus stuff for next year....

myininaya (myininaya):

i don't see how to do this it failed at integral by parts

OpenStudy (anonymous):

I'm working on it....gimme a sec

myininaya (myininaya):

\[\int\limits_{}^{}\frac{e^{\frac{x}{2}}}{\sqrt{x}} dx\] ? right?

OpenStudy (anonymous):

ya! that's the question

OpenStudy (anonymous):

and part 2 of the question is find the derivative

myininaya (myininaya):

i still say there is no elementary way to evaluate this i would like someone to evaluate it, but im not sure how many people on here no complex analysis or whatever else you need for this

OpenStudy (anonymous):

oh... thanks for working on this problem anyway :)

myininaya (myininaya):

blue hows it going?

OpenStudy (anonymous):

Hmm..not great, you may be right...I'll let you know if I have a breakthrough.

myininaya (myininaya):

lol gl i think i will leave this site up all night and check it randomly to see if anyone answers this

OpenStudy (anonymous):

Blue were you trying to use integration by parts

OpenStudy (anonymous):

You're right, you can't do it. I put it into wolfram. One of the components of the answer is the erf function.

OpenStudy (anonymous):

Yes zeal, first by parts and then by u substitution with u=sqrt(x), then x=u^2, etc...but then you just end up with e^(0.5u^2) in the integrand and it gets you nowhere. I honestly find it kind of hard to believe that they'd give you this, or any other, integral in a precalc class.

OpenStudy (anonymous):

thank you guys for working on this problem...! btw, is finding the derivative for y=(e^(x/2)) / (sqrtx) also impossible too?

OpenStudy (anonymous):

No, finding the derivative of that is pretty straightforward...just gotta follow some rules (chain, quotient, power, etc).

OpenStudy (anonymous):

i substituted sqrt of x first... is that correct?

OpenStudy (anonymous):

Hmm? No substituting for derivatives. Have you learned the quotient or chain rules yet?

OpenStudy (anonymous):

ya... i got [(e^(x/2))*(sqrt of x)]/2 + [(e^(x/2)) / (2*sqrt of x}] as the answer.... is that right?

myininaya (myininaya):

(e^(x/2))'=1/2*e^(x/2) (x^(1/2))'=1/(2sqr{x}) so we need to use quotient rule since we have f/g quotient rule: (f/g)'=(f'g-fg')/g^2 =(1/2*e^(x/2)*x^(1/2)-e^(x/2)*1/(2sqrt{x}))/x and we can make this look nice by getting rid of the compund fraction

myininaya (myininaya):

multply top and bottom by 2sqrt{x} and that will do the trick i have to go to bed good night

OpenStudy (anonymous):

thanks!

OpenStudy (anonymous):

oh i forgot about that g^2

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