If f(x) = 2tan⁻¹x + sin⁻¹(2x+x²), x > 1, then f(5) is equal to:

f(x) = 2tan⁻¹x + sin⁻¹(2x + x²)
Let x = 5
f(5) = 2tan⁻¹(5) + sin⁻¹(10 + 25) = 2tan⁻¹(5) + sin⁻¹(35)
Since sin⁻¹(35) is not defined (as the range of sin⁻¹(x) is [-π/2, π/2] and 35 >1), we must use the property sin⁻¹(x) + cos⁻¹(x) = π/2
Let's use another approach:
Let y = tan⁻¹x
tan y = x
sin y = x/√(1+x²)
cos y = 1/√(1+x²)
2tan⁻¹x = 2y
Consider sin⁻¹(2x + x²) = sin⁻¹(2x + x²) = θ
sin θ = 2x + x²
cos θ = √(1 - (2x + x²)²)
2tan⁻¹x + sin⁻¹(2x + x²) = 2y + θ
If x = 5, then
2tan⁻¹(5) + sin⁻¹(35)
This is not defined since the range of sin⁻¹(x) is [-π/2, π/2]
Let's assume there's a mistake in the question and the term is sin⁻¹(2x/(1+x²))
Then, sin⁻¹(2x/(1+x²)) = 2tan⁻¹x
f(x) = 2tan⁻¹x + 2tan⁻¹x = 4tan⁻¹x
f(5) = 4tan⁻¹(5)