The value of the definite integral _ -1 ^1 ( ⁻¹ x)^2 1+e^x dx is

Step-by-step solution

Let I = _ -1 ^1 ( ⁻¹ x)^2 1+e^x dx Using the property _ -a ^a f(x) dx = ₀^a (f(x) + f(-x)) dx , we get: I = ₀^1 ( ( ⁻¹ x)^2 1+e^x + ( ⁻¹(-x))^2 1+e^ -x ) dx Since ⁻¹(-x) = - ⁻¹ x , we have ( ⁻¹(-x))^2 = (- ⁻¹ x)^2 = ( ⁻¹ x)^2 . I = ₀^1 ( ⁻¹ x)^2 ( 1 1+e^x + e^x e^x+1 ) dx I = ₀^1 ( ⁻¹ x)^2 dx Substitute x = dx = d . When x = 0, = 0 ; when x = 1, = 2 . I = ₀^ /2 ^2 d Using integration by parts: I = [ ^2 ]₀^ /2 - ₀^ /2 2 d I = ^2 4 - 2 [ - + ]₀^ /2 I = ^2 4 - 2(1 - 0) = ^2 4 - 2 . Answer: ^2 4 - 2