integrate [sec inverse of (x+1/x-1)] + [sin inverse of (x-1/x+1)

We need to evaluate the integral:

∫ [sec⁻¹((x + 1) / (x - 1))] + [sin⁻¹((x - 1) / (x + 1))] dx

Step 1: Define Functions
Let:
I = ∫ [sec⁻¹((x + 1) / (x - 1))] + [sin⁻¹((x - 1) / (x + 1))] dx

Step 2: Check for Possible Simplifications
We analyze the two inverse trigonometric functions separately.

Expression 1: sec⁻¹((x + 1) / (x - 1))
For the given function sec⁻¹(y), we know that:
sec⁻¹(y) + sin⁻¹(1/y) = π/2, for y > 1

If we set y = (x + 1) / (x - 1), then its reciprocal is:
1/y = (x - 1) / (x + 1)

Thus,
sin⁻¹((x - 1) / (x + 1)) = π/2 - sec⁻¹((x + 1) / (x - 1))

Step 3: Substituting the Identity
From the identity above, we substitute in our integral:

I = ∫ [sec⁻¹((x + 1) / (x - 1))] + [π/2 - sec⁻¹((x + 1) / (x - 1))] dx

This simplifies to:
I = ∫ [π/2] dx

Step 4: Final Integration
Since π/2 is a constant, its integral is straightforward:

I = (π/2) ∫ dx
I = (π/2) x + C

Final Answer:
I = (π/2) x + C