Which of the following is a correct antiderivative form for ∫ sqrt(1+4x^2) dx used in arc length?

• A. (x/2) sqrt(1+4x^2) + (1/2) asinh(2x) + C
• B. (x/2) sqrt(1+4x^2) + (1/4) asinh(2x) + C
• C. (x/2) sqrt(1+4x^2) - (1/4) asinh(2x) + C
• D. (x/2) sqrt(1+4x^2) + asinh(2x) + C

Answer: (B)

Think about turning this into a standard form you recognize. Notice sqrt(1+4x^2) is sqrt(1+(2x)^2). Let u = 2x, so du = 2 dx and dx = du/2. The integral becomes (1/2) ∫ sqrt(1+u^2) du, which uses the known antiderivative ∫ sqrt(1+u^2) du = (u/2) sqrt(1+u^2) + (1/2) asinh(u) + C. Multiply by 1/2 to get (u/4) sqrt(1+u^2) + (1/4) asinh(u) + C. Substituting back u = 2x gives (x/2) sqrt(1+4x^2) + (1/4) asinh(2x) + C. This is the correct antiderivative for arc length in this case. The coefficient on the asinh term must be 1/4 after the substitution; other given forms would not differentiate back to sqrt(1+4x^2).