Consider f(x) = \frac{x^2}{1+x^3}; g(t) = \int f(t)dt. If g(1) = 0 then g(x) equals -
\frac{1}{3} \ln(1+x^3)
\frac{1}{3} \ln\left(\frac{1+x^3}{2}\right)
\frac{1}{2} \ln\left(\frac{1+x^3}{3}\right)
\frac{1}{3} \ln\left(\frac{1+x^3}{3}\right)
Step-by-Step Solution
Key Concept: The integral of f(x) = x^2/(1+x^3) is found by substitution u = 1+x^3, du = 3x^2 dx. Thus, g(x) = (1/3)ln|1+x^3| + C. Using g(1) = 0, we find C = -(1/3)ln(2).
Given f(x) = x^2 / (1+x^3). Let g(x) = \int (x^2 / (1+x^3)) dx. Let u = 1+x^3, then du = 3x^2 dx, so x^2 dx = du/3. Thus, g(x) = (1/3) \int (1/u) du = (1/3) ln|1+x^3| + C. Given g(1) = 0, we have (1/3) ln(1+1^3) + C = 0, which means (1/3) ln(2) + C = 0, so C = -(1/3) ln(2). Therefore, g(x) = (1/3) ln(1+x^3) - (1/3) ln(2) = (1/3) ln((1+x^3)/2).
Correct Answer: 2