_ 0^ 2 x 1+ x+ x d x=

Step-by-step solution

Let I = ₀^ /2 x 1 + x + x dx ... (1) Using ₀^a f(x)dx = ₀^a f(a-x)dx , substitute x 2 - x : I = ₀^ /2 x 1 + x + x dx ... (2) Adding (1) and (2): 2I = ₀^ /2 x + x 1 + x + x dx = ₀^ /2 (1 - 1 1 + x + x )dx 2I = 2 - J where J = ₀^ /2 1 1 + x + x dx Using half-angle: x = 2 ^2 x 2 - 1 , x = 2 x 2 x 2 : J = ₀^ /2 1 2 ^2 x 2 + 2 x 2 x 2 dx Dividing by 2 ^2 x 2 : J = ₀^ /2 1 2 ^2 x 2 1 + x 2 dx Let t = 1 + x 2 , dt = 1 2 ^2 x 2 dx . Limits: t: 1 2 J = ₁^2 1 t dt = 2 2I = 2 - 2 I = 4 - 1 2 2