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Argonx16
 2 years ago
Best ResponseYou've already chosen the best response.0Integrate: \[\int\limits_{1}^{3} \frac {1}{x+1} dx \] I'm brain farting hard right now. Probably only need like 2 steps worked out.

Argonx16
 2 years ago
Best ResponseYou've already chosen the best response.0EDIT: I totally meant \[\int\limits_{1}^{3} \frac {x}{x+1}dx\]

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.2It's a simple Usubstitution with a little trick involved. It's very easy to get confused on a problem like this one. :)

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.1You have to do a little trick with usubstitution in this case.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.2Let \(\large u=x+1\), solve for x giving us \(\large x=u1\). Then differentiating gives us, \(\large du=dx\).

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.1I guess I will let Zepdrix finish it off :P .

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.2Understand what's going on argon? :O It's just that little piece \(\large x=u1\) that is easy to forget about.

Argonx16
 2 years ago
Best ResponseYou've already chosen the best response.0@Dido525 Thanks anyways c: @zepdrix Ok, let me go through it, one sec.

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.1If you want I can show you another way to do this question other than what Zepdrix is doing.

Argonx16
 2 years ago
Best ResponseYou've already chosen the best response.0@zepdrix So I looked at my work, and I have tried usubstitution, and I did get the fact that du=dx (u=x+1, du=1 dx, du=dx), but how does that fit into the big picture? @Dido525 Thanks, but I'm fine for now :D

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.2So what will happen is, you'll have the addition/subtraction on TOP instead of the bottom of the fraction when you perform this substitution. That's good news, cause we'll be able to break it up into a couple of fractions from there.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.2Before we had that messy term x+1 in the bottom, so we couldn't do much with it.. but with a u1 in the top, we can split it up nicely.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.2\[\large \int\limits \frac{x}{x+1}dx \qquad \rightarrow \qquad \int\limits \frac{u1}{u}du\]I think we get something like this yes? Which we can write as,\[\large \int\limits \frac{u}{u}\frac{1}{u}du\]

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.1Long division would yield the same thing by the way. Just a note.

Argonx16
 2 years ago
Best ResponseYou've already chosen the best response.0Wow... that's some clever manipulation! (I GET IT!) @Dido525 Now that you mention it, I think my teacher did say that long division was necessary on some of the questions.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.2Yah that makes more sense actually, instead of doing a substitution. Just add and subtract 1 from the top.\[\large \int\limits \frac{x+11}{x+1}dx \qquad \rightarrow \qquad \int\limits \frac{x+1}{x+1}\frac{1}{x+1}dx\]

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.1Well it's not NECESSARY but it's DEFINITELY Easier :P .

Argonx16
 2 years ago
Best ResponseYou've already chosen the best response.0Ok. Thanks so much guys! I was so puzzled as to why Wolfram Alpha had \[\int\limits_ {}^{} 1 \frac{1}{x+1}\]
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