$\quad \int 3^{3^x} \cdot 3^x d x=$

Question

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Let } \mathrm{I}=\int 3^{3^x} \cdot 3^x \mathrm{~d} x \ & 	ext { Let } 3^x=\mathrm{t} \ & 3^x \log 3 \mathrm{~d} x=\mathrm{dt} \ & 3^x \mathrm{~d} x=\frac{1}{\log 3} \mathrm{dt} \ & 	herefore \quad \mathrm{I}=\int 3^{\mathrm{t}} \cdot \frac{1}{\log 3} \cdot \mathrm{dt} \ & \quad=\frac{1}{\log 3} \int 3^{\mathrm{t}} \mathrm{dt} \ & \quad=\frac{1}{\log 3} 	imes \frac{3^{\mathrm{t}}}{\log 3}+\mathrm{c} \ & \mathrm{I}=\frac{3^{3^x}}{(\log 3)^2}+\mathrm{c}\end{aligned}$

$\quad \int 3^{3^x} \cdot 3^x d x=$

Solution