int x^2 (3)^{x^3 + 1} dx = | vedclass

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(D) Let $I = \int x^2 (3)^{x^3 + 1} dx$. We know that $3^{x^3 + 1} = 3 \cdot 3^{x^3}$. So,$I = \int x^2 \cdot 3 \cdot 3^{x^3} dx = 3 \int x^2 \cdot 3^

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(D) Let $I = \int x^2 (3)^{x^3 + 1} dx$. We know that $3^{x^3 + 1} = 3 \cdot 3^{x^3}$. So,$I = \int x^2 \cdot 3 \cdot 3^{x^3} dx = 3 \int x^2 \cdot 3^{x^3} dx$. Let $t = x^3$. Then $dt = 3x^2 dx$,which implies $x^2 dx = \frac{dt}{3}$. Substituting these into the integral: $I = 3 \int 3^t \cdot \frac{dt}{3} = \int 3^t dt$. Using the formula $\int a^t dt = \frac{a^t}{\ln a} + c$,we get: $I = \frac{3^t}{\ln 3} + c$. Substituting $t = x^3$ back,we get: $I = \frac{3^{x^3}}{\ln 3} + c$. Since the original expression was $3^{x^3+1}$,note that $\frac{3^{x^3+1}}{3 \ln 3} = \frac{3 \cdot 3^{x^3}}{3 \ln 3} = \frac{3^{x^3}}{\ln 3}$. Thus,the correct form is $\frac{3^{x^3}}{\ln 3} + c$.

$\int x^2 (3)^{x^3 + 1} dx = $

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