## anonymous 3 years ago Integration (See inside)

1. anonymous

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.

2. anonymous

EDIT: I totally meant $\int\limits_{1}^{3} \frac {x}{x+1}dx$

3. zepdrix

It's a simple U-substitution with a little trick involved. It's very easy to get confused on a problem like this one. :)

4. anonymous

You have to do a little trick with u-substitution in this case.

5. zepdrix

Let $$\large u=x+1$$, solve for x giving us $$\large x=u-1$$. Then differentiating gives us, $$\large du=dx$$.

6. anonymous

I guess I will let Zepdrix finish it off :P .

7. zepdrix

Understand what's going on argon? :O It's just that little piece $$\large x=u-1$$ that is easy to forget about.

8. anonymous

@Dido525 Thanks anyways c: @zepdrix Ok, let me go through it, one sec.

9. anonymous

If you want I can show you another way to do this question other than what Zepdrix is doing.

10. anonymous

@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

11. zepdrix

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.

12. zepdrix

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.

13. zepdrix

$\large \int\limits \frac{x}{x+1}dx \qquad \rightarrow \qquad \int\limits \frac{u-1}{u}du$I think we get something like this yes? Which we can write as,$\large \int\limits \frac{u}{u}-\frac{1}{u}du$

14. anonymous

Long division would yield the same thing by the way. Just a note.

15. zepdrix

Ah good call :3

16. anonymous

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.

17. zepdrix

Yah that makes more sense actually, instead of doing a substitution. Just add and subtract 1 from the top.$\large \int\limits \frac{x+1-1}{x+1}dx \qquad \rightarrow \qquad \int\limits \frac{x+1}{x+1}-\frac{1}{x+1}dx$

18. anonymous

Well it's not NECESSARY but it's DEFINITELY Easier :P .

19. anonymous

Ok. Thanks so much guys! I was so puzzled as to why Wolfram Alpha had $\int\limits_ {}^{} 1- \frac{1}{x+1}$

20. anonymous

Long division again.

21. anonymous

Yep!