∫777x⋅77x⋅7xdx is equal to

7 7 7 x ⋅ ( log ⁡ 7 ) 3 + C

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$\int 7^{7^{7^{x}}} \cdot 7^{7^{x}} \cdot 7^{x} d x$ is equal to

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$\frac{7^{7^{7^{x}}}}{{(\log_{e} 7)}^{3}} + C$

none of these

$7^{7^{7^{x}}} \cdot (\log 7)^{3} + C$

$\frac{7^{7^{7^{x}}}}{{(\log_{e} 7)}^{2}} + C$

answer is A.

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Detailed Solution

Let

$\begin{array}{l} I = \int 7^{7^{7^{x}}} \cdot 7^{7^{x}} \cdot 7^{x} d x \\ \Rightarrow I = \int \frac{1}{{(\log_{e} 7)}^{3}} \int (7^{7^{7^{x}}} \log_{e} 7) \cdot (7^{7^{x}} \log_{e} 7) \cdot (7^{x} \log_{e} 7) d x \\ \Rightarrow I = \frac{1}{{(\log_{e} 7)}^{3}} \int 1 \cdot d (7^{7^{x}}) = \frac{7^{7^{x^{x}}}}{{(\log_{e} 7)}^{3}} + C \end{array}$