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Argonx16
Integration (See inside)
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.
EDIT: I totally meant \[\int\limits_{1}^{3} \frac {x}{x+1}dx\]
It's a simple U-substitution with a little trick involved. It's very easy to get confused on a problem like this one. :)
You have to do a little trick with u-substitution in this case.
Let \(\large u=x+1\), solve for x giving us \(\large x=u-1\). Then differentiating gives us, \(\large du=dx\).
I guess I will let Zepdrix finish it off :P .
Understand what's going on argon? :O It's just that little piece \(\large x=u-1\) that is easy to forget about.
@Dido525 Thanks anyways c: @zepdrix Ok, let me go through it, one sec.
If you want I can show you another way to do this question other than what Zepdrix is doing.
@zepdrix So I looked at my work, and I have tried u-substitution, 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
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.
Before we had that messy term x+1 in the bottom, so we couldn't do much with it.. but with a u-1 in the top, we can split it up nicely.
\[\large \int\limits \frac{x}{x+1}dx \qquad \rightarrow \qquad \int\limits \frac{u-1}{u}du\]I think we get something like this yes? Which we can write as,\[\large \int\limits \frac{u}{u}-\frac{1}{u}du\]
Long division would yield the same thing by the way. Just a note.
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.
Yah that makes more sense actually, instead of doing a substitution. Just add and subtract 1 from the top.\[\large \int\limits \frac{x+1-1}{x+1}dx \qquad \rightarrow \qquad \int\limits \frac{x+1}{x+1}-\frac{1}{x+1}dx\]
Well it's not NECESSARY but it's DEFINITELY Easier :P .
Ok. Thanks so much guys! I was so puzzled as to why Wolfram Alpha had \[\int\limits_ {}^{} 1- \frac{1}{x+1}\]