The value of ₀^1 ⁻¹ (1-x+x^2 ) d x is

Step-by-step solution

The integral I = ₀^1 ⁻¹(1-x+x^2) dx can be evaluated using trigonometric identities and substitution techniques. Since 1-x+x^2 = (x- 1 2 )^2 + 3 4 > 0 for x [0,1] , the identity ⁻¹ A + ⁻¹ 1 A = 2 applies, giving ⁻¹(1-x+x^2) = 2 - ⁻¹ ( 1 1-x+x^2 ) . Substituting yields I = ₀^1 2 dx - ₀^1 ⁻¹ ( 1 1-x+x^2 ) dx = 2 - ₀^1 ⁻¹ ( 1 1-x+x^2 ) dx . Using the identity ⁻¹ x - ⁻¹ (x-1) = ⁻¹ ( 1 1-x+x^2 ) , the integral becomes I = 2 - ₀^1 ( ⁻¹ x - ⁻¹ (x-1)) dx . Substituting u = x-1 in the second term gives ₀^1 ⁻¹ (x-1) dx = _ -