int 3^{3^x} cdot 3^x , dx = | vedclass

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Question

(C) Let $I = \int 3^{3^x} \cdot 3^x \, dx$. Substitute $t = 3^x$. Then,$dt = 3^x \log 3 \, dx$,which implies $3^x \, dx = \frac{1}{\log 3} \, dt$. Sub

Vedclass · Accepted Answer

$\frac{3^{3^x}}{(\log 3)^2} + c$,where $c$ is a constant of integration.

Vedclass · Answer

(C) Let $I = \int 3^{3^x} \cdot 3^x \, dx$. Substitute $t = 3^x$. Then,$dt = 3^x \log 3 \, dx$,which implies $3^x \, dx = \frac{1}{\log 3} \, dt$. Substituting these into the integral: $I = \int 3^t \cdot \frac{1}{\log 3} \, dt$. $I = \frac{1}{\log 3} \int 3^t \, dt$. Using the formula $\int a^x \, dx = \frac{a^x}{\log a} + c$: $I = \frac{1}{\log 3} \cdot \frac{3^t}{\log 3} + c$. $I = \frac{3^t}{(\log 3)^2} + c$. Substituting $t = 3^x$ back: $I = \frac{3^{3^x}}{(\log 3)^2} + c$.

$\int 3^{3^x} \cdot 3^x \, dx =$

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