If ∫ (1/x+1/x³)(²³√(3x⁻²⁴+x⁻²⁶))dx=-α/3(α+1)(3x^β+x^γ)^((α+1)/α)+C,x>0,(α,β,γ∈ Z), where C is the constant of integration, then α+β+γ is equal to ___________.

Solution

Maths sir · 4 steps

Pull x⁻²³ out of the root first; then the bracket left inside is exactly what you substitute.

1. Take out a power. Write 3x⁻²⁴+x⁻²⁶ = x⁻²³(3x⁻¹+x⁻³). The 23rd root of x⁻²³ is x⁻¹, because x>0.
2. Rewrite the integrand. Multiply that x⁻¹ into the first bracket. The integral becomes ∫ (1/x²+1/x⁴)(3x⁻¹+x⁻³)^(1/23)dx.
3. Substitute. Put t = 3x⁻¹+x⁻³. Then dt = (-3/x²-3/x⁴)dx = -3(1/x²+1/x⁴)dx, which matches the first bracket.

   -1/3∫ t^(1/23) dt = -1/3·23/24 t^(24/23) + C = -23/3(24)(3x⁻¹+x⁻³)^(24/23) + C

4. So. Matching with the given form, α = 23, β = -1 and γ = -3. Hence α+β+γ = 23-1-3 = 19.

The common slip: 27. Dropping the negative signs and taking β = 1, γ = 3 gives 23+1+3 = 27, not 19. The bracket is 3x⁻¹+x⁻³.

The idea to keep. When a root holds a sum of powers of x, factor out the power that the root removes exactly. The bracket that is left is usually the substitution, and its derivative sits outside.

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