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JEE Main202523 Jan 2025Morning ShiftMathematicsDefinite IntegrationActual

The value of _ e^2 ^ e^4 1 x ( e^ ( ( _e x )^2+1 )⁻¹ e^ ( ( _e x )^2+1 )⁻¹ +e^ ( (6- _e x )^2+1 )⁻¹ ) d x is

Options

  1. A2
  2. B_e 2
  3. C1
  4. De^2

Correct answer

C. 1

Step-by-step solution

aligned & Put x=t 1 x d x=d t array |c|c| x & t e^2 & 2 e^4 & 4 array & I= ₂^4 e^ ( t ^2+1 )⁻¹ e^ ( t ^2+1 )⁻¹ +e^ ((6-t)^2+1 )⁻¹ d t (i) & I= ₂^4 e^ ((6-t)^2+1 )⁻¹ e^ ((6-t)^2+1 )⁻¹ +e^ (t^2+1 )⁻¹ d t (ii) & Using _a^0 f(x) d x= _a^0 f(a+b-x) d x aligned Adding (i) and (ii) gives 2 l= d t l=1

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