Question

Note: This is NOT a tutorial. This is a question.

\[\LARGE \int _{-1}^{0} \frac{e^{\frac{1}{x}}}{x^3} dx\]

please show me the steps :D

note: will no longer be making tutorials for an indefinite time due to some legal issues. i will return to doing them once the issues are fixed. thank you

Answer 1

APPLY ILATE rule

Answer 2

please start from BEGINNING :p

Answer 3

take
\[ 1/x = t\]
\[\frac{-dx}{x^2}=dt\]
Write x =1/t
\[{dx}=\frac{-dt}{t^2}\]

Changing the limit, we get
Lower limit t =1/-1 = -1
Upper limit t = 1/0 = infinity
Substituting this, we get
\[\large \int\limits_{-1}^{\infty} \frac{e^t}{\frac{1}{t^3}}{(\frac{-dt}{t^2})}\]
\[\large \int\limits_{-1}^{\infty}- e^t tdt\]
Now apply udv rule

Take u = t
dv = e^t dt

find du, v from above

Now according to udv rule
\[\int\limits_{}^{}u dv = u \int\limits_{}^{} dv - v \int\limits_{}^{}du \]
Substitute and get your answer :)

Answer 4

@lgbasallote , is it clear ?? :D

Answer 5

uhmm ill be afk for a while..ill check it later and ask questions :)

Answer 6

ugh i dont get it from beginninng sorry @shivam_bhalla i think it's me :/ i'll just keep practicing