Consider f(x) = frac{x^2}{1 + x^3} and g(t) = | vedclass

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(B) Given $f(x) = \frac{x^2}{1 + x^3}$.We need to find $g(x) = \int f(x) \, dx = \int \frac{x^2}{1 + x^3} \, dx$.Let $u = 1 + x^3$. Then $du = 3x^2 \,

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$\frac{1}{3} \ln\left( \frac{1 + x^3}{2} \right)$

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(B) Given $f(x) = \frac{x^2}{1 + x^3}$.We need to find $g(x) = \int f(x) \, dx = \int \frac{x^2}{1 + x^3} \, dx$.Let $u = 1 + x^3$. Then $du = 3x^2 \, dx$,which implies $x^2 \, dx = \frac{du}{3}$.Substituting these into the integral:$g(x) = \int \frac{1}{u} \cdot \frac{du}{3} = \frac{1}{3} \ln|u| + C = \frac{1}{3} \ln(1 + x^3) + C$.Given the condition $g(1) = 0$:$0 = \frac{1}{3} \ln(1 + 1^3) + C \implies 0 = \frac{1}{3} \ln(2) + C \implies C = -\frac{1}{3} \ln(2)$.Substituting $C$ back into the expression for $g(x)$:$g(x) = \frac{1}{3} \ln(1 + x^3) - \frac{1}{3} \ln(2) = \frac{1}{3} \ln\left( \frac{1 + x^3}{2} \right)$.

Consider $f(x) = \frac{x^2}{1 + x^3}$ and $g(t) = \int f(t) \, dt$. If $g(1) = 0$,then $g(x)$ equals:

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