Let I(x) = e^ e^x + 2x (e^x+1)^2 dx and I(0) = e 2 . Then the value of I( 2) is equal to

Step-by-step solution

Given, I(x) = e^ e^x + 2x (e^x+1)^2 dx I(x) = e^ e^x e^ 2x (e^x+1)^2 dx Let e^x = t e^x dx = dt I(x) = e^t t (t+1)^2 dt I(x) = e^t ( t+1-1 (t+1)^2 ) dt I(x) = e^t ( 1 t+1 - 1 (t+1)^2 ) dt This is of the form e^t (f(t) + f'(t)) dt = e^t f(t) + C , where f(t) = 1 t+1 and f'(t) = - 1 (t+1)^2 . I(x) = e^t t+1 + C = e^ e^x e^x+1 + C Now, using I(0) = e 2 : I(0) = e^ e^0 e^0+1 + C = e 2 + C e 2 + C = e 2 C = 0 Hence, the value of I( 2) is: I( 2) = e^ e^ 2 e^ 2 +1 = e^2 2+1 = e^2 3 Answer: e^2 3