Integrate ∫sqrt((x+1)/(x-1))

In summary, the conversation discusses the integration of the function ∫sqrt((x+1)/(x-1)), with the attempt at a solution involving substitution and simplification. The suggested substitution of x = cosh(t) is also mentioned.
  • #1
Alex235123
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0

Homework Statement


∫sqrt((x+1)/(x-1))

Homework Equations

The Attempt at a Solution


t=sqrt((x+1)/(x-1)), t^2=(x+1)/(x-1)⇒x=(-t^2-1)/(-t^2+1) dx=dt⇒ -4t/((t^2-1)^2)
∫t*-4t/((t^2-1)^2)=-4∫t^2+1-1/((t^2+1)^2)=-4∫dt/t^2+1-4∫dt/((t^2-1)^2)
 
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  • #2
Please post calculus HW problems in the Calculus HW forum.
 
  • #3
Alex235123 said:

Homework Statement


∫sqrt((x+1)/(x-1))

Homework Equations

The Attempt at a Solution


t=sqrt((x+1)/(x-1)), t^2=(x+1)/(x-1)⇒x=(-t^2-1)/(-t^2+1) dx=dt⇒ -4t/((t^2-1)^2)
∫t*-4t/((t^2-1)^2)=-4∫t^2+1-1/((t^2+1)^2)=-4∫dt/t^2+1-4∫dt/((t^2-1)^2)
Hello Alex235123. Welcome to PF !

It's really not proper to leave the dx out of the integral. It's especially important when using substitution.

Consider multiplying the numerator and denominator by x+1, then simplifying the integrand.

The substitution x = cosh(t) looks like it works well.
 

1. How do I approach integrating this function?

To integrate this function, you can use the substitution method by substituting u = √((x+1)/(x-1)) and simplifying the expression to make it easier to integrate.

2. Can this function be integrated using any other method?

Yes, you can also integrate this function using the partial fractions method by rewriting the function as a sum of simpler fractions.

3. Is there a specific domain for this function?

Yes, the function is only defined for x ≠ 1, since the denominator becomes zero at x = 1.

4. Can this function be integrated by hand, or do I need a calculator?

This function can be integrated by hand, although it may be a bit tedious. Using a calculator can make the process quicker and more accurate.

5. What are some real-world applications of this function?

This type of function can be used in physics to calculate the work done by a force on an object, as well as in economics to calculate the area under a demand or supply curve.

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