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Argonx16Best ResponseYou've already chosen the best response.0
Integrate: \[\int\limits_{1}^{3} \frac {1}{x+1} dx \] I'm brain farting hard right now. Probably only need like 2 steps worked out.
 one year ago

Argonx16Best ResponseYou've already chosen the best response.0
EDIT: I totally meant \[\int\limits_{1}^{3} \frac {x}{x+1}dx\]
 one year ago

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

Dido525Best ResponseYou've already chosen the best response.1
You have to do a little trick with usubstitution in this case.
 one year ago

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

Dido525Best ResponseYou've already chosen the best response.1
I guess I will let Zepdrix finish it off :P .
 one year ago

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

Argonx16Best ResponseYou've already chosen the best response.0
@Dido525 Thanks anyways c: @zepdrix Ok, let me go through it, one sec.
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
If you want I can show you another way to do this question other than what Zepdrix is doing.
 one year ago

Argonx16Best 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
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
So 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.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
Before 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.
 one year ago

zepdrixBest 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,\[\large \int\limits \frac{u}{u}\frac{1}{u}du\]
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
Long division would yield the same thing by the way. Just a note.
 one year ago

Argonx16Best ResponseYou've already chosen the best response.0
Wow... 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.
 one year ago

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

Dido525Best ResponseYou've already chosen the best response.1
Well it's not NECESSARY but it's DEFINITELY Easier :P .
 one year ago

Argonx16Best ResponseYou've already chosen the best response.0
Ok. Thanks so much guys! I was so puzzled as to why Wolfram Alpha had \[\int\limits_ {}^{} 1 \frac{1}{x+1}\]
 one year ago
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