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Mathematics 34 Online
OpenStudy (anonymous):

Another quirky problem.. Find the indefinite integral: Integrand of (e^(-1/x))/x^2 Would i just subtract the two, or use substitution?

OpenStudy (anonymous):

or rather, just find the integral (not indefinite)

OpenStudy (lgbasallote):

\[\huge \int \frac{e^{\frac{-1}{x}}}{x^2}dx?\]

OpenStudy (anonymous):

yes

OpenStudy (lgbasallote):

let u = -1/x that'll do the trick...do a u-substitution

OpenStudy (anonymous):

would du= lnx?

OpenStudy (lgbasallote):

nope...derivative of -1/x is not lnx

OpenStudy (anonymous):

oh sorry it would be x^-2?

OpenStudy (lgbasallote):

yup!!

OpenStudy (anonymous):

so then w/ substitution it would be: (e^u/x^2) du?

OpenStudy (anonymous):

or is it du on the denominator?

OpenStudy (lgbasallote):

can i ask why there is an x^2 in the denominator?

OpenStudy (anonymous):

that's what I'm confused about, when i substitute what happens to the denominator of the original equation?

OpenStudy (lgbasallote):

\[\huge \int \frac{e^{-\frac{1}{x}}}{x^2}dx = \int (e^{-\frac{1}{x}})(\frac{1}{x^2})dx\] do you agree?

OpenStudy (anonymous):

yes

OpenStudy (lgbasallote):

u = -1/x du = (1/x^2)dx

OpenStudy (anonymous):

ok, yup i got that.. now how do I solve

OpenStudy (lgbasallote):

\[\LARGE \int e^u du\] can you see that?

OpenStudy (anonymous):

yeah.. and that's the part i question because it looks like the answer would be just the same as the original?

OpenStudy (lgbasallote):

not the original...the x^2 in the denominator will be annihilated after substitution

OpenStudy (lgbasallote):

\[\LARGE \int e^u du = e^u = e^{-\frac{1}{x}}\]

OpenStudy (anonymous):

oh this is where du is left off.. hmm ok i think i understand

OpenStudy (lgbasallote):

great :D

OpenStudy (anonymous):

thank you!

OpenStudy (anonymous):

I may have other questions :)

OpenStudy (lgbasallote):

\[\huge \color{maroon}{\mathbb{\text{<tips hat>}}}\] just post it as another question

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