∫333x33x3xdx is equal to

$\int 3^{3^{3^{x}}} 3^{3^{x}} 3^{x} d x$ is equal to

A
$\frac{x^{3^{x}}}{(\log 3)^{3}} + C$
B
$3^{3^{3^{x}}} (\log 3)^{3} + C$
C
$\frac{3^{3^{3^{x}}}}{(\log 3)^{3}} + C$
D
none of these

Solution:

Put $3^{3^{3^{x}}} = t \Rightarrow (\log 3)^{3} 3^{3^{3^{x}}} \cdot 3^{3^{x}} \cdot 3 x = \frac{d t}{d x}$

So given integral is given to $\int \frac{d t}{(\log 3)^{3}} = \frac{t}{(\log 3)^{3}} + C .$

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