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int 1/x^2 e^(-1/x) dx=...

`int 1/x^2 e^(-1/x) dx=`

A

`e^(-1//x)+C`

B

`-e^(-1//x)+C`

C

`(e^(-1//x))/(x)+C`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
A
Put `-(1)/(x)=t and(1)/(x^(2))dx=dt`.
`:." "I=inte^(t)dt=e^(t)+C=e^(-1//x)+C`.
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