For α,β,γ,δ∈ ℕ, if ∫ ((x/e)²ˣ+(e/x)²ˣ)logₑ xdx=1/α(x/e)^(β x)-1/γ(e/x)^(δ x)+C , where e=Σₙ₌₀^∞1/n! and C is constant of integration, then α+2β+3γ-4δ is equal to :

Solution

Maths sir · 4 steps

Do not integrate head-on: differentiate (x/e)²ˣ and watch logₑ x appear on its own.

1. Take logs first. Let u = (x/e)²ˣ. Then ln u = 2x(ln x - 1), so u'/u = 2(ln x - 1) + 2x · 1/x = 2 ln x.
2. Same for the other. Let v = (e/x)²ˣ. Then ln v = 2x(1 - ln x), so v'/v = 2(1 - ln x) - 2 = -2 ln x.

   d/dx[1/2(x/e)²ˣ - 1/2(e/x)²ˣ] = ((x/e)²ˣ + (e/x)²ˣ)logₑ x

3. Read off values. So the integral is 1/2(x/e)²ˣ - 1/2(e/x)²ˣ + C, which gives α = β = γ = δ = 2.
4. So. α + 2β + 3γ - 4δ = 2 + 4 + 6 - 8 = 4, which is option (d).

The idea to keep. When the integrand has xˣ-type powers multiplied by ln x, differentiate the power using logs. The ln x factor is usually its derivative in disguise.

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