How do I integrate this? x/(x^2-1)^.5

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Discussion Overview

The discussion revolves around the integration of the expressions 1/(x^2-1)^(1/2) and x/(x^2-1)^(1/2). Participants explore various methods for solving these integrals, including u-substitution and trigonometric substitution, while addressing potential complications in the integration process.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant suggests using u-substitution for the integral of 1/(x^2-1)^(1/2).
  • Another participant proposes a trigonometric substitution, specifically suggesting x = cosh(y) for the integral of 1/(x^2-1)^(1/2), but notes a discrepancy in the expected answer.
  • A different approach is introduced using the substitution x = sec(θ), leading to the integral being expressed in terms of secant and tangent functions.
  • One participant points out a potential oversight in simplifying the integral of sinh(y)/sinh(y) to 1 + constant.
  • A final suggestion involves a u-substitution where u = x^2 - 1, indicating a need for adjustment in the differential.

Areas of Agreement / Disagreement

Participants present multiple competing views on how to approach the integration problem, with no consensus reached on the best method or the correctness of the derived answers.

Contextual Notes

There are unresolved issues regarding the correctness of the transformations and substitutions used in the integration process, as well as the expected outcomes of the integrals.

ComFlu945
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How do I integrate this? 1/(x^2-1)^.5

How do I integrate this? x/(x^2-1)^.5

And this
1/(x^2-1)^.5
 
Last edited:
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For the first, make a u-substitution.

For the second, make a trigonometric substitution suggested by that difference of squares.
 
Thanks!

First one worked like a charm.

For second one I substituted x for cosh(y). Since cosh(y)^2-1=sinh(y)^2, but bottom turns into sinh(y). And since x=cosh(y), dx/dy= sinh(y).

Back to original equation:
integral( 1/(x^2-1)^.5 dx) = integral ( sinh(y)/sinh(y)) dy = 1 + constant. However, the answer is supposed to be cosh^-1(x).
 
Let x=secø
dx=secøtanødø
tanø=√(x2-1)

So your integral becomes:
∫dx/√(x2-1) = ∫secødø = ln|secø+tanø|

Substituting back in
ln|secø+tanø|=ln|x+√(x2-1)|
 
ComFlu945 said:
integral ( sinh(y)/sinh(y)) dy = 1 + constant

Think about that some more
 
By u substitution, let u=x^2-1 then du=2x
you have x you only need 2
 

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