Question

evaluate integral 0 to 1 (x)/(x^2 - 1)

Answer 1

This requires u-substitution. Arre you familiar with this method?

Answer 2

\[\int\limits_{0}^{1}\frac{ x }{ x^2-1 }dx\]

Answer 3

no

Answer 4

\(u = x^2-1\) while \(du=2x\) you don't have a 2x, so you're going ot have to divide both sides by 2. You want something that you can substitute. You have \[\int\limits_{0}^{1}\frac{ 2 }{ u }du = 2 \int\limits \frac{ 1 }{ u }du\] now clearly, you can do the integral of that, which is simply the natural log plus a constant. but you have an definite integral, so you don't need a constant.
Don't for get to substitute back your original "u" which was \(u=x^2-1\) REMEMBER.
some people like to change the bounds and evaluate using "u", but i find prefer to evaluate using original function, since I don't have to change anything.
So byu fundamental theorem of calculus II
you should know that you have \(F(b)-F(a)\) where you're bound is 0<x<1 or [0,1]

Answer 5

Okay but this integral does not converge :/

Answer 6

Oh, you're right. What am I doing. Lol

Answer 7

lol

Answer 8

so? should i just say that or?

Answer 9

Do you know WHY it doesn't converge?first of all. :)

Answer 10

ohhh lol

Answer 11

As you can see, your problem is the upper limit.

Answer 12

okay

Answer 13

so, evaluate the integral from 0 to some variable, say "R"....and take the limit.

Answer 14

umm ok

Answer 15

Lol, where are you confused?

Answer 16

so its convegent?

Answer 17

@RadEn can u help?

Answer 18

the integral not exist for upper interval x = 1