The function y=f(x) is the solution of the differential equation dydx + xy/(x²⁻¹)=x⁴+2x√(1-x²) in (-1,1) satisfying f(0)=0. Then ∫⁻√3/2√3/2 f(x)dx is

dy/dx + xy/(x²-1) = x⁴+2x√(1-x²)
This is a linear differential equation
I.F. = e∫xy/(x²-1)dx = e^(1/2ln|x²-1|) = √(1-x²)
⇒solution is y√(1-x²) = ∫x(x³+2)√(1-x²)⋅√(1-x²)dx
or y√(1-x²) = ∫(x⁴+2x)dx = x⁵/5 + x² + c
f(0) = 0 ⇒ c = 0 ⇒ f(x)√(1-x²) = x⁵/5 + x²
Now, ∫⁻√3/2√3/2 f(x)dx = ∫⁻√3/2√3/2 (x⁵/5 + x²)/√(1-x²) dx (Using property) = 2∫⁰√3/2 (x⁵/5 + x²)/√(1-x²) dx = 2∫π/3⁰ sin²θcosθ/cosθ dθ (Taking x = sinθ) = 2∫π/3⁰ sin²θ dθ = 2[θ/2 - sin2θ/4]π/3⁰ = 2(π/6) - (√3/8) = π/3 - √3/4