Question:

Step 1: Define the given function $f(x)$ and set up the integral for $g(x)$. We are given the function $f(x) = \frac{x^2}{1+x^3}$ and $g(t) = \int f(t)dt$. Therefore, we can write $g(x)$ as: $$g(x) = \int \frac{x^2}{1+x^3} dx$$ Step 2: Perform the integration using a substitution method. To integrate, we use the substitution method. Let $u = 1+x^3$. Differentiating both sides with respect to $x$, we get: $$\frac{du}{dx} = 3x^2$$ $$du = 3x^2 dx$$ From this, we can express $x^2 dx$ as $\frac{1}{3} du$. Now, substitute $u$ and $x^2 dx$ into the integral for $g(x)$: $$g(x) = \int \frac{1}{u} \cdot \frac{1}{3} du$$ $$g(x) = \frac{1}{3} \int \frac{1}{u} du$$ Integrating $\frac{1}{u}$ with respect to $u$ gives $\ln|u|$: $$g(x) = \frac{1}{3} \ln|u| + C$$ Substitute back $u = 1+x^3$: $$g(x) = \frac{1}{3} \ln|1+x^3| + C$$ Since $1+x^3$ is positive for the relevant domain implied by the options, we can drop the absolute value: $$g(x) = \frac{1}{3} \ln(1+x^3) + C$$ Step 3: Use the initial condition $g(1) = 0$ to find the constant of integration $C$. We are given that $g(1)=0$. Substitute $x=1$ into the expression for $g(x)$: $$g(1) = \frac{1}{3} \ln(1+1^3) + C = 0$$ $$\frac{1}{3} \ln(1+1) + C = 0$$ $$\frac{1}{3} \ln(2) + C = 0$$ Solving for $C$: $$C = -\frac{1}{3} \ln(2)$$ Step 4: Substitute the constant $C$ back into $g(x)$ and simplify the expression. Substitute the value of $C$ back into the equation for $g(x)$: $$g(x) = \frac{1}{3} \ln(1+x^3) - \frac{1}{3} \ln(2)$$ Factor out $\frac{1}{3}$: $$g(x) = \frac{1}{3} \left( \ln(1+x^3) - \ln(2) \right)$$ Using the logarithm property $\ln A - \ln B = \ln \left(\frac{A}{B}\right)$: $$g(x) = \frac{1}{3} \ln\left(\frac{1+x^3}{2}\right)$$ Step 5: State the final answer and match it with the correct option. The function $g(x)$ is $\frac{1}{3} \ln\left(\frac{1+x^3}{2}\right)$. Comparing this with the given options, it matches Option 2. The final answer is $\boxed{\text{Option 2}}$.