₀^ 5 e^x e^x - 1 e^x + 3 ,dx =

Step-by-step solution

Let e^x - 1 = t^2 . Differentiating both sides, we get e^x dx = 2t dt . When x = 0 , t^2 = e^0 - 1 = 0 t = 0 . When x = 5 , t^2 = e^ 5 - 1 = 5 - 1 = 4 t = 2 . The denominator becomes e^x + 3 = (t^2 + 1) + 3 = t^2 + 4 . Substituting these into the integral, we get: ₀^ 5 e^x - 1 e^x + 3 (e^x dx) = ₀^2 t t^2 + 4 (2t dt) = ₀^2 2t^2 t^2 + 4 dt = 2 ₀^2 t^2 + 4 - 4 t^2 + 4 dt = 2 ₀^2 ( 1 - 4 t^2 + 4 ) dt = 2 [ t - 4 ( 1 2 ⁻¹ ( t 2 ) ) ]₀^2 = 2 [ t - 2 ⁻¹ ( t 2 ) ]₀^2 = 2 [ ( 2 - 2 ⁻¹(1) ) - (0 - 0) ] = 2 ( 2 - 2 ( 4 ) ) =