Question

Find the integral e^(3/x)/x^2

Answer 1

Have you considered a substitution?

Answer 2

im really confused with u substitution

Answer 3

im thinking that in this question: u=x^2 and du=2x dx right?

Answer 4

how do you know which to substitute?

Answer 5

Try u = 1/x. This leads to du = -1/x^2 dx and vastly simplifies things.

You must think of some significant part of the argument that will be the derivative of your substitution. In this case, I noticed that we have (1/x) in that exponent and the derivative of 1/x is -1/x^2 which just happens to match the denominator. Perfect!

Answer 6

where are you gettin 1/x though?

Answer 7

you mean 3/x?

Answer 8

No. 3/x = 3(1/x) You can use u 3/x if you like, you'll just get more constants to fix. If the problem were \(3\cdot\dfrac{e^{3/x}}{x^{2}}\), I would have picked u = 3/x.

Note: It won't always be obvious. You have to think about it and try something. You will gain experience as you go along.

Note: There may be more than one way to proceed. You do not have to find THE WAY. Find a way that works. Various constants don't matter. You can fix those on the fly.

Answer 9

* u = 3/x if you like

Answer 10

if u=3/x then du=-3x^-2

Answer 11

right?

Answer 12

\(du = -\dfrac{3}{x^{2}}dx\)

Answer 13

so then\[\int\limits\frac{ e^u }{ x^2 } dx?\]

Answer 14

Well, sort of. You must do the rest. Substitute for \(\dfrac{dx}{x^{2}}\).

Answer 15

Given \(du = -\dfrac{3}{x^{2}}dx\), we have \(\dfrac{dx}{x^{2}} = \dfrac{du}{-3}\)

Answer 16

and you take the integral of that? sorry im just really confused

Answer 17

Substitute ALL the pieces. The point is to make an integral that you can handle.

We have: \(\int\dfrac{e^{3/x}}{x^{2}}\;dx\)

We decide \(u = \dfrac{3}{x}\)

We calculate: \(du = -\dfrac{3}{x^{2}}dx\)

We rebuild the integral so the substitutionS will work perfectly \(\dfrac{1}{-3}\cdot\int\dfrac{e^{3/x}}{x^{2}}\cdot \left(-3\right)\;dx\)

We substitute EVERTHING: \(\dfrac{1}{-3}\cdot\int\e^{u}\;du\)

We solve the simpler integral: \(\dfrac{1}{-3}e^{u} + C\)

We put things back: \(\dfrac{1}{-3}e^{3/x} + C\)

Done.

Answer 18

sorry to keep asking...but im still wondering how you got 1/-3? in the front of the integral

Answer 19

I just put it on the outside because I needed one on the inside.

\(\int Stuff\;dx = \dfrac{1}{a}\int a\cdot Stuff\;dx\)

I could have written: \(\int \dfrac{a}{a}\cdot Stuff \;dx\)

You should know this principle: \(\int a\cdot f(x)\;dx = a\cdot\int f(x)\;dx\)

Answer 20

ok that makes much more sense

Answer 21

thanks for all your help and patience tkhunny :)

Answer 22

I needed WAY more partience for the connection problems than for you.

Way to hang in there. Good work.

Answer 23

thanks!