If ∫₀^(√3) (15 x³)/√(1+x²+√(1+x²)³) d x=α √2+β √3, where α, β are integers, then α+β is equal to __________.

Solution

Maths sir · 4 steps

Put u = √(1+x²) and the messy root becomes u√(1+u), so the integral turns into a polynomial.

1. Substitute. Let u = √(1+x²), so 1+x² = u² and x dx = u du. The denominator becomes √(u² + u³) = u√(1+u). When x goes from 0 to √3, u goes from 1 to 2.
2. Simplify. Write x³ dx = x² · x dx = (u²-1) u du. The u cancels, and (u²-1)/√(1+u) = (u-1)√(1+u). Now put v = 1+u, which runs from 2 to 3.

   15∫₂³(v^(3/2) - 2v^(1/2))dv = [ 6v^(5/2) - 20v^(3/2) ]₂³

3. Evaluate. At v = 3: 54√3 - 60√3 = -6√3. At v = 2: 24√2 - 40√2 = -16√2. Upper minus lower gives 16√2 - 6√3.
4. So. α = 16 and β = -6, so α + β = 10.

The common slip: -10. Subtracting upper minus lower the wrong way round gives α = -16, β = 6 and a sum of -10. Always take the value at the upper limit minus the value at the lower limit.

The idea to keep. When √(1+x²) shows up with powers, substitute u = √(1+x²) and use x dx = u du. Change the limits too, so you never have to go back to x.

Answer from the NTA final answer key. Solution by PrepWiser.