Question

Find thw anti-derivative of
\[f(x)=\frac{ -1 }{ x^2 }e ^{1/x}\]

Answer 1

\[\large -\int \frac{e^{1/x}}{x^2}\]

Make a u-sub of 1/x

Answer 2

Oh...what have you learned?

Answer 3

How do you write in this format?
f(x)=F(x)+C

Answer 4

\[\Large\rm \int\limits f(x)dx=F(x)+C\]Where F(x) is the anti-derivative of f(x).
Did you understand the u-substitution?

Answer 5

Is there another way to solve it without using the u-substitution?

Answer 6

i can't think of a better way

Answer 7

The reason that I ask that was because the question was at the beginning of antiderivatives/integration. I learned the u-substitution as a last topic...

Answer 8

you may use advanced guessing
what is the derivative of \(\large e^{1/x}\) ?

Answer 9

e^1/x

Answer 10

what do you know about chain rule

Answer 11

Chain rule applies to composite function. It describe of outer evulated at inner, times derivative of inner

Answer 12

So using that logic...if we consider the "inner" as 1/x

The derivative of \(\large e^{1/x}\) would be \(\large \frac{d(1/x)}{dx} \times e^{1/x}\)

So what is the derivative of \(\large 1/x\) ?

Answer 13

lnx?

Answer 14

you found the antiderivative for positive x

Answer 15

that you found the antiderivative of 1/x for positive x
the question was to differentiate

Answer 16

derivative if lnx is 1/x

derivative of 1/x is not lnx

Answer 17

ooh... derivative of 1/x is -1/x^2 ...

Answer 18

Correct...and to finish up from the previous post

\[\large \frac{d}{dx}e^{1/x} = e^{1/x} \times \frac{d}{dx}\frac{1}{x}\]

Since we just found the later \(\large \frac{d}{dx} \frac{1}{x} = -\frac{1}{x^2}\) we have

\[\large \frac{d}{dx}e^{1/x} = -\frac{e^{1/x}}{x^2}\] right?

Now what do you notice?

Answer 19

So F(x)= e^{1/x} +C ?

Answer 20

Perfect! :)

Answer 21

Thank you! :) I got it now. I was messed up b/w derivatives and anti-derivatives. My bad...

Answer 22

Yeah sometimes they can be hard to keep straight...but as long as you can derive them you're all set!