Another quirky problem.. Find the indefinite integral: Integrand of (e^(-1/x))/x^2 Would i just subtract the two, or use substitution?
or rather, just find the integral (not indefinite)
\[\huge \int \frac{e^{\frac{-1}{x}}}{x^2}dx?\]
yes
let u = -1/x that'll do the trick...do a u-substitution
would du= lnx?
nope...derivative of -1/x is not lnx
oh sorry it would be x^-2?
yup!!
so then w/ substitution it would be: (e^u/x^2) du?
or is it du on the denominator?
can i ask why there is an x^2 in the denominator?
that's what I'm confused about, when i substitute what happens to the denominator of the original equation?
\[\huge \int \frac{e^{-\frac{1}{x}}}{x^2}dx = \int (e^{-\frac{1}{x}})(\frac{1}{x^2})dx\] do you agree?
yes
u = -1/x du = (1/x^2)dx
ok, yup i got that.. now how do I solve
\[\LARGE \int e^u du\] can you see that?
yeah.. and that's the part i question because it looks like the answer would be just the same as the original?
not the original...the x^2 in the denominator will be annihilated after substitution
\[\LARGE \int e^u du = e^u = e^{-\frac{1}{x}}\]
oh this is where du is left off.. hmm ok i think i understand
great :D
thank you!
I may have other questions :)
\[\huge \color{maroon}{\mathbb{\text{<tips hat>}}}\] just post it as another question
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