If ∫₀¹ 1/(5+2 x-2 x²)(1+e⁽²⁻⁴ˣ⁾) d x=1/α log ₑ((α+1)/β), α, β>0, then α⁴-β⁴ is equal to :

Solution

Maths sir · 4 steps

Use x → 1-x: the quadratic stays the same, the exponential flips, and adding the two kills it.

1. Reflect the variable. Under x → 1-x, 5+2x-2x² stays exactly the same, and e²⁻⁴ˣ becomes e⁻⁽²⁻⁴ˣ⁾. Write t = 2-4x. Since 1/(1+eᵗ)+1/(1+e⁻ᵗ) = 1, adding the original integral to the new one removes the exponential.

   2I = ∫₀¹ dx/(5+2x-2x²) = ∫₀¹ dx/2(11/4-(x-1/2)²)

2. Evaluate. Put u = x-1/2, with a = √11/2, and use ∫ du/(a²-u²) = 1/2a ln(a+u)/(a-u) from u=-1/2 to 1/2. This gives 2I = 1/√11 ln(√11+1)/(√11-1).
3. Match the form. Rationalise: (√11+1)/(√11-1) = (√11+1)²/10. So I = 1/2√11 ln(√11+1)²/10 = 1/√11 ln(√11+1)/√10. Comparing with 1/α ln(α+1)/β gives α = √11 and β = √10.
4. So. α⁴-β⁴ = 121-100 = 21, which is option (b).

The idea to keep. When you see 1/(1+eᶠ⁽ˣ⁾) and x → a+b-x turns f into -f, add the two forms of the integral. The exponential part sums to 1 and disappears.

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