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

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x3x(log⁡3)3+C

333x(log⁡3)3+C

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detailed solution

Correct option is C

Put 333x=t⇒(log⁡3)3333x⋅33x⋅3x=dtdxSo given integral is given to ∫dt(log⁡3)3=t(log⁡3)3+C.

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∫333x33x3xdx is equal to

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