If the value of the integral _ -1 ¹ ( x ⁻¹ x 1+e^x + ⁻¹ x 1+x^2 ) dx is expressed in the form 8 (a + b ) , where a and b are integers, then the value of a + b is

Step-by-step solution

Let I = _ -1 ¹ ( x ⁻¹ x 1+e^x + ⁻¹ x 1+x^2 ) dx = I₁ + I₂ . For I₁ = _ -1 ¹ x ⁻¹ x 1+e^x dx , we use the property _ -A ^ A f(x) dx = ₀^ A (f(x) + f(-x)) dx . Since x ⁻¹ x is an even function, f(x) + f(-x) = x ⁻¹ x 1+e^x + (-x) ⁻¹ (-x) 1+e^ -x = x ⁻¹ x ( 1 1+e^x + e^x 1+e^x ) = x ⁻¹ x . So, I₁ = ₀¹ x ⁻¹ x dx . Using integration by parts for I₁ : I₁ = [ x^2 2 ⁻¹ x ]₀¹ - ₀¹ x^2 2 1-x^2 dx = 1 2 2 - 1 2 ₀¹ x^2 1-x^2 dx Substitute x = dx = d in the remaining integral: ₀¹ x^2 1-x^2 dx = ₀^ 2 ^2 d = ₀^ 2 ^2 d = 4 . Thus,