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Mathematics 7 Online
OpenStudy (anonymous):

Integrate...

OpenStudy (anonymous):

\[\int\limits_{}^{}\frac{ 1 }{ (\ln s) +4 }ds\]

OpenStudy (anonymous):

is there a way I'm forgetting that can simplify the denominator

hartnn (hartnn):

u can't simplify the denominator further.

OpenStudy (anonymous):

u substitution I don't think will work

hartnn (hartnn):

nopes, this integration is not standard.... answer involves exponential integral http://www.wolframalpha.com/input/?i=integral+1%2F%284%2Bln+x%29

OpenStudy (amistre64):

u = lns + 4 ; s= e^(u-4) du = 1/s ds s du = ds \[\int \frac{e^{u-4}}{u}du\] it does seem a bit odd eh

OpenStudy (amistre64):

we might be able to change that into a suitable taylor poly tho

OpenStudy (anonymous):

I see what you did. Very creative nice.

OpenStudy (amistre64):

thnx :) not sure how useful it is tho ....

OpenStudy (anonymous):

It is pretty messy. The real question was is this equation separable and thats what i got on one side. I was just curious what the integration would look like, but I didn't have to solve

OpenStudy (amistre64):

just a bit more mashing it about \[\int \frac{e^{u-4}}{u}du\] \[\int \frac{e^{u}e^{-4}}{u}du\] \[e^{-4}\int \frac{e^{u}}{u}du\]

OpenStudy (anonymous):

and no closed form for what amistre got

OpenStudy (anonymous):

I could use an explanation on how do identify a linear equation if you have the time. I have read the definition in the book and it is just not clicking

OpenStudy (amistre64):

i have a test to take in 15 minutes, and a 10 minutes walk to get there :) good luck

OpenStudy (anonymous):

A linear first order equation is an equation that can be expressed in the form\[a _{1}(x)\frac{ dy }{ dx }+a _{0}(x)y= b(x)\]

OpenStudy (anonymous):

Thanks good luck to you too.

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