The value of ∫_(e²)^(e⁴) 1/x(e^(((log ₑ x)²+1)⁻¹)/(e^(((log ₑ x)²+1)⁻¹)+e^(((6-log ₑ x)²+1)⁻¹))) d x is

Solution

Maths sir · 4 steps

Set t = ln x: the limits 2 and 4 add to 6, the same 6 in the bracket, so use King's property.

1. Change the variable. Let t = ln x, so dt = 1/x dx. The limits x = e² and x = e⁴ become t = 2 and t = 4. Write f(t) = e^(1/(t²+1)) to keep things short.

   I = ∫₂⁴ f(t)/(f(t) + f(6-t)) dt

2. Apply King's property. Use ∫ₐᵇ g(t) dt = ∫ₐᵇ g(a+b-t) dt with a + b = 6. Replacing t by 6 - t swaps f(t) and f(6-t), so I = ∫₂⁴ f(6-t)/(f(6-t) + f(t)) dt.
3. Add the two forms. The two numerators add up to the common denominator, so 2I = ∫₂⁴ 1 dt = 4 - 2 = 2.
4. So. I = 1, which is option (a).

The idea to keep. When the integrand looks like f(t)/(f(t)+f(a+b-t)), King's property makes the two forms add to 1, so the integral is half the length of the interval, (b-a)/2.

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