The value of the definite integral ₀^ _e 5 e^x e^x-1 e^x+3 ,dx is equal to...

Step-by-step solution

Let e^x - 1 = t^2 , which gives e^x dx = 2t dt . When x = 0 , t = 0 . When x = _e 5 , t = 2 . The integral becomes: I = ₀^2 t t^2 + 1 + 3 (2t dt) = ₀^2 2t^2 t^2 + 4 dt I = 2 ₀^2 t^2 + 4 - 4 t^2 + 4 dt I = 2 ₀^2 ( 1 - 4 t^2 + 4 ) dt I = 2 [ t - 4 1 2 ⁻¹ ( t 2 ) ]₀^2 I = 2 [ 2 - 2 ⁻¹(1) ] I = 2 ( 2 - 2 4 ) I = 4 - Answer: 4 -