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: Integrate f(x) = x^2/(1+x^3) using substitution u = 1+x^3, then use the condition g(1)=0 to find the constant of integration.
Given f(x) = \frac{x^2}{1+x^3}. Then g(x) = \int \frac{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) = \frac{1}{3} \int \frac{du}{u} = \frac{1}{3} \ln|1+x^3| + C. Given g(1) = 0, we have \frac{1}{3} \ln(1+1^3) + C = 0, which means \frac{1}{3} \ln(2) + C = 0, so C = -\frac{1}{3} \ln(2). Therefore, g(x) = \frac{1}{3} \ln(1+x^3) - \frac{1}{3} \ln(2) = \frac{1}{3} \ln\left(\frac{1+x^3}{2}\right).
Correct Answer: B