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## 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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- Thread starter Alex235123
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∫sqrt((x+1)/(x-1))

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)

- #2

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Please post calculus HW problems in the Calculus HW forum.

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Hello Alex235123. Welcome to PF !## 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)

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.

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