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Simple Integration using U Substitution

by anon413
Tags: integration, simple, substitution
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anon413
#1
Jan28-08, 01:42 PM
P: 13
1. The problem statement, all variables and given/known data
Find the indefinite integral.
The antiderivative or the integral of (x^2-1)/(x^2-1)^(1/2)dx



2. Relevant equations



3. The attempt at a solution

Tried using (x^2-1)^(1/2) as u and udu for dx and I solved for x but I am still left with a 1 on top not sure how to deal with it.
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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rocomath
#2
Jan28-08, 02:00 PM
rocomath's Avatar
P: 1,754
[tex]\int\frac{x^2-1}{\sqrt{2x-1}}dx[/tex]

Yes?
anon413
#3
Jan28-08, 02:05 PM
P: 13
Sorry made a mistake in typing the denominator (2x-1)^(1/2) Im slightly displexic

anon413
#4
Jan28-08, 02:05 PM
P: 13
Simple Integration using U Substitution

How did you type that exactly.
rocomath
#5
Jan28-08, 02:08 PM
rocomath's Avatar
P: 1,754
Quote Quote by anon413 View Post
How did you type that exactly.
LaTeX http://physicsforums.com/showthread.php?t=8997

Check my thread again, I editted it. Is it correct now?
anon413
#6
Jan28-08, 02:12 PM
P: 13
Yes thank you, that command latex is complicated I'll try to learn it.
sutupidmath
#7
Jan28-08, 02:13 PM
P: 1,633
well try to use this substitution:
2x-1=u^2, try to defferentiate and substitute back what u get for dx,
you also get for x=(u^2 + 1)/2, from 2x-1=u^2
the rest is pretty simple after u substitute!
Can you go from here?
rocomath
#8
Jan28-08, 02:15 PM
rocomath's Avatar
P: 1,754
Quote Quote by anon413 View Post
Yes thank you, that command latex is complicated I'll try to learn it.
Gets easier as you use it on a daily basis.

First, I would break up the numerator.

[tex]\int\frac{x^2}{\sqrt{2x-1}}dx-\int\frac{dx}{\sqrt{2x-1}}[/tex]
sutupidmath
#9
Jan28-08, 02:17 PM
P: 1,633
Quote Quote by rocophysics View Post
Gets easier as you use it on a daily basis.

First, I would break up the numerator.

[tex]\int\frac{x^2}{\sqrt{2x-1}}dx-\int\frac{dx}{\sqrt{2x-1}}[/tex]
It might work this way also, but by immediately taking the substitution that i suggested he will get to the result pretty fast.
anon413
#10
Jan28-08, 02:22 PM
P: 13
Ok ill try stupid maths method I would get the integral of (((u^2+1)/2)^2-1)/(u) what would my du be
anon413
#11
Jan28-08, 02:43 PM
P: 13
I guess I will attempt this on my own thats for the help.
sutupidmath
#12
Jan28-08, 03:57 PM
P: 1,633
You do not need to know what the du will be, you need just to plug in the value of the dx that you get after differentiating 2x-1=u^2, so it obviously will be dx=udu, and when you plug this in the u here and that one that you will get on the denominator will cancel out so you are left with somehting like this:
[tex]\int\frac{((\frac{u^{2}+1}{2})^{2}-1)udu}{u}[/tex], now you can go from here right?
sutupidmath
#13
Jan28-08, 04:02 PM
P: 1,633
Quote Quote by anon413 View Post
Ok ill try stupid maths method I would get the integral of (((u^2+1)/2)^2-1)/(u) what would my du be
By the way it is not stupid math, but instead sutupidmath! NOt that i mind it, but it feels good to be correct! NO hard feelings, ok?
anon413
#14
Jan28-08, 05:15 PM
P: 13
thx stupidmath. I knew I got the right udu for dx.


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