₀^1 d x e^x+e^ -x =

Step-by-step solution

Let I = ₀¹ dx e^x + e^ -x . Multiply the numerator and denominator by e^x : I = ₀¹ e^x dx e^ 2x + 1 . Substitute u = e^x , so du = e^x dx . When x=0 , u=1 . When x=1 , u=e . I = ₁^ e du u^2 + 1 . The integral of 1 u^2 + 1 is ⁻¹(u) . I = [ ⁻¹(u)]₁^ e = ⁻¹(e) - ⁻¹(1) . Since ⁻¹(1) = 4 , we have I = ⁻¹(e) - 4 . Answer: ⁻¹ e- 4