Let I(x) = x^2 - x x^2 + e^ 2x dx , x > 0 . If _ x I(x) = 1 , then I(1) is equal to

Step-by-step solution

Given I(x) = x^2 - x x^2 + e^ 2x dx Dividing the numerator and the denominator by x^2 , we get: I(x) = 1 - 1 x 1 + ( e^x x )^2 dx Let t = e^x x Differentiating both sides with respect to x : dt = x e^x - e^x x^2 dx = e^x x (1 - 1 x ) dx = t (1 - 1 x ) dx (1 - 1 x ) dx = dt t Substituting this into the integral: I(x) = 1 1 + t^2 dt t = 1 t(1 + t^2) dt Using partial fractions: I(x) = ( 1 t - t 1 + t^2 ) dt I(x) = |t| - 1 2 (1 + t^2) + C = ( t 1 + t^2 ) + C Substituting back t = e^x x : I(x) = ( e^x x 1 + ( e^x x )^2