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COMEDK20269 May 2026Morning ShiftMathematicsIndefinite IntegrationActual

e^ (1+ 1 x^2 ) x^2 + 1 x^2 dx =

Options

  1. A1 2 ⁻¹ ( x^2-1 2 ,x ) + C
  2. B1 2 ⁻¹ (x - 1 x ) + C
  3. C1 2 ⁻¹ ( x^2+1 x 2 ) + C
  4. D- 1 2 ⁻¹ (x - 1 x ) + C

Correct answer

A. 1 2 ⁻¹ ( x^2-1 2 ,x ) + C

Step-by-step solution

Given integral is I = e^ (1+ 1 x^2 ) x^2 + 1 x^2 dx Using the property e^ f(x) = f(x) , the integral simplifies to: I = 1+ 1 x^2 x^2 + 1 x^2 dx The denominator can be rewritten by completing the square: x^2 + 1 x^2 = (x - 1 x )^2 + 2 So the integral becomes: I = 1+ 1 x^2 (x - 1 x )^2 + 2 dx Let x - 1 x = t Differentiating both sides with respect to x , we get: (1 + 1 x^2 ) dx = dt Substituting this into the integral: I = dt t^2 + ( 2 )^2 Using the standard integral formula dx x^2 + a^2 = 1 a ⁻¹ ( x a ) + C : I = 1

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