U substitution. Why -1/x^2 is my du?

[PauloE]

Homework Statement

∫(1/x^(2))(3+1/x)^(3)

Homework Equations

U substitution is the way to go here

The Attempt at a Solution

My problem is that I can't figure my du and what is next. I know which one it is but I don't know the reason for it.

u=3+1/x
du= I chose ln|x| first but that got me nowhere. My buddy told me (txt) me it is -1/x^(2) which eventually takes you to -(u)^4/4 but i can't figure out what happened in between.

thanks for the hand.

 

[PauloE]

BTW i just need someone to explain me how to choose du. I'm fine after that

 

[SammyS]

Homework Statement

∫(1/x^(2))(3+1/x)^(3)

Homework Equations

U substitution is the way to go here

The Attempt at a Solution

My problem is that I can't figure my du and what is next. I know which one it is but I don't know the reason for it.

u=3+1/x
du= I chose ln|x| first but that got me nowhere. My buddy told me (txt) me it is -1/x^(2) which eventually takes you to -(u)^4/4 but i can't figure out what happened in between.

thanks for the hand.

(Use the superscript, X², icon for exponents.)
First of all, it's a big help to include the dx with your integral:

∫(1/x²)(3+1/x)³ dx

It can look even better in LaTeX .

$\displaystyle \int \frac{1}{x^2}\left(3+\frac{1}{x}\right)^3\, dx$

What is the derivative of $\displaystyle \left(3+\frac{1}{x}\right)^3 \ ?$

(That won't give the answer directly, but it may lead you to it.)
.

 

[PauloE]

alright, with the chain rule I get 3(3+1/x)^2*-1/x^2

-1/x^2 would be my du then. (I was deriving it wrong before) so now that I have this minus sign I put it outside the Integrate sign. after that I'm good. Thanks!

So if instead of a -1 i would have a 3 do I deal with it as if it were k and put it outside of the integral?

BTW I've just downloaded LaTex

 

[Dick]

alright, with the chain rule I get 3(3+1/x)^2*-1/x^2

-1/x^2 would be my du then. (I was deriving it wrong before) so now that I have this minus sign I put it outside the Integrate sign. after that I'm good. Thanks!

So if instead of a -1 i would have a 3 do I deal with it as if it were k and put it outside of the integral?

BTW I've just downloaded LaTex

You can always move a constant outside of an integral. And you didn't need to download LaTex. It's built into this site. Try looking here https://www.physicsforums.com/showpost.php?p=3977517&postcount=3

 

[PauloE]

[itex]∫(1/x^2)(3+1/x)^3[\itex] test

 

[Dick]

[itex]∫(1/x^2)(3+1/x)^3[\itex] test

$∫(1/x^2)(3+1/x)^3$ The slash in front of \itex is in the wrong direction. It should be /itex.

 

[SammyS]

alright, with the chain rule I get 3(3+1/x)^2*-1/x^2

-1/x^2 would be my du then. (I was deriving it wrong before) so now that I have this minus sign I put it outside the Integrate sign. after that I'm good. Thanks!

So if instead of a -1 I would have a 3 do I deal with it as if it were k and put it outside of the integral?

BTW I've just downloaded LaTex

You don't seem to like dx !

du would be -1/x²dx , multiplied by some constant.

Since you end up with (3+1/x)² , and you need (3+1/x)^3, you must need to start with a different power on the (3 + 1/x) factor.

 

[PauloE]

You don't seem to like dx !

du would be -1/x²dx , multiplied by some constant.

Since you end up with (3+1/x)² , and you need (3+1/x)^3, you must need to start with a different power on the (3 + 1/x) factor.

true, I have to pay more attention with writing dx

My final answer is:

[itex]-((3+1/x)⁴/4) + c[/itex]

thanks for the help!

 

[Mark44]

true, I have to pay more attention with writing dx

My final answer is:

[itex]-((3+1/x)⁴/4) + c[/itex]

thanks for the help!

You can't mix LaTeX with the <sup> tags. Here's what you wrote, fixed.
$-((3 + 1/x)^4/4) + C$

My LaTeX script looks like this:
-((3 + 1/x)^4/4) + C

 

[Dick]

true, I have to pay more attention with writing dx

My final answer is:

[itex]-((3+1/x)⁴/4) + c[/itex]

thanks for the help!

Right, $-((3+1/x)^4/4) + c$. If you are wondering why it didn't TeX, don't use the shortcut superscripts and stuff inside Tex. Use ^ to get superscripts.

 

[PauloE]

$-((3 + 1/x)^4/4) + C$ let's see now

Thank y'all for the help