Primitive !

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

\[\int\limits_{}^{}e^\frac{ -1 }{ t }\]
dt
|dw:1436757696187:dw|

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

\(\large\color{black}{ \displaystyle \int e^{-1/t}~dt }\) \(\large\color{black}{ \displaystyle u=-1/t }\) \(\large\color{black}{ \displaystyle du=1/t^2~dt }\) \(\large\color{black}{ \displaystyle t^2~du=dt }\) \(\large\color{black}{ \displaystyle (-1/u)^2~du=dt }\) \(\large\color{black}{ \displaystyle 1/u^2~du=dt }\) \(\large\color{black}{ \displaystyle \int \frac{e^{u}}{u^2}~du }\) I don't think this has an answer in term of a simple function. What I see to do instead is: \(\large\color{black}{ \displaystyle e^t=\sum_{n=0}^{\infty}\frac{t^n}{n!} }\) \(\large\color{black}{ \displaystyle e^{-1/t}=\sum_{n=0}^{\infty}\frac{(-1/t)^n}{n!} }\) \(\large\color{black}{ \displaystyle e^{-1/t}=\sum_{n=0}^{\infty}\frac{(-1)^nt^{-n}}{n!} }\) \(\large\color{black}{ \displaystyle \int e^{-1/t}~dt~=\int \sum_{n=0}^{\infty}\frac{(-1)^nt^{-n}}{n!} dt }\) \(\large\color{black}{ \displaystyle \int e^{-1/t}~dt~=\sum_{n=0}^{\infty}\frac{(-1)^nt^{-n+1}}{~(-n+1)~n!~} +C }\) and that you can simplify as: \(\large\color{black}{ \displaystyle \int e^{-1/t}~dt~=\sum_{n=0}^{\infty}\frac{(-1)^n}{~~t^{n-1}~(-n+1)~n!~} +C }\)
I mean if you haven't learned this technique using a series, then disregard my reply....
I agree that there's no antiderivative in terms of elementary functions. This seems relevant: http://mathworld.wolfram.com/ExponentialIntegral.html You should be able to get that if you integrate by parts. (I haven't bothered checking for myself, though.)
Oh, I have seen a post with this idea of integration by parts a certain number of times after which you arrive at an infinite series. I have done a similar thing when integrated e^x/x. (of course differentiating the 1/x) This is how I got to know the series technique, but without proving/showing using by parts every time is more convinient:D

Not the answer you are looking for?

Search for more explanations.

Ask your own question