Mark M. answered 03/26/21
Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
Let u = 1+x2. So, du = 2xdx. Therefore, xdx = (1/2)du.
∫x√(1+x2)dx = ∫ u1/2(1/2)du = (1/2)(2/3)u3/2 + C = (1/3)(1 + x2)3/2 + C
Tori P.
asked 03/26/21Mark M. answered 03/26/21
Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
Let u = 1+x2. So, du = 2xdx. Therefore, xdx = (1/2)du.
∫x√(1+x2)dx = ∫ u1/2(1/2)du = (1/2)(2/3)u3/2 + C = (1/3)(1 + x2)3/2 + C
Raymond B. answered 03/26/21
Math, microeconomics or criminal justice
let u = 1+x^2 then du = 2x
integral of u^1/2 du = u^3/2/3/2 + a constant
= (2/3)u^3/2 + a constant
= (2/3)(1+ x^2)^3/2 + a constant
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