Calculus

We can factor this expression using the difference of squares formula, which states that for any x, (x^2-a^2)=(x-a)(x+a). In this case, we can replace a with 1 to get (x^2-1)=(x-1)(x+1): ((x^2+1)^(1/2)*(x^2-1)+((x^2+1)^(3)^(1/2)*(2x^2-2)) =((x^2+1)^(1/2)*(x-1)(x+1)+((x^2+1)^(3)^(1/2)*(2x^2-2)) =((x^2+1)^(1/2)*(x+1)(x-1)+((x^2+1)^(3)^(1/2)*2(x^2-1)) =(x^2+1)^(1/2)*(x+1)(x-1)+2((x^2+1)^(3/2)*(x^2-1)) The expression can now be factored as: (x^2+1)^(1/2)*(x+1)(x-1)+2((x^2+1)