\(\int 7^{7^{7^{x}}} 7^{7^{x}} 7^{x} \mathrm{~d} x=\)

Let I \(=\int 7^{7^{7^{x}}} 7^{7^{x}} 7^{x} \mathrm{dx}\)
Let \(y=7^{7^{7^{x}}}\)
\(\log y=7^{7^{x}}(\log 7)\)
Differentiating w.r.t. x, we get
\(
\frac{1}{y} \frac{d y}{d x}=(\log 7) \frac{d}{d x}\left(7^{7^{x}}\right)
\)
Let \(z=7^{7^{x}} \Rightarrow \log z=7^{x} \log 7\)
Differentiating w.r.t. \(\mathrm{x}\), we get
\(\frac{1}{z} \frac{d z}{d x}=7^{x}(\log 7)^{2} \)
\( \therefore \frac{d}{d x}\left(7^{7^{x}}\right)=7^{7^{x}} 7^{x}(\log 7)^{2} \Rightarrow \frac{d y}{d x}=\)\(7^{7^{7}} 7^{7^{x}} 7^{x}(\log 7)^{3} \)
\( \therefore I=\int \frac{d}{d x}\left(7^{7^{7^{x}}}\right) \times \frac{1}{(\log 7)^{3}} d x=\frac{7^{7^{x}}}{(\log 7)^{3}}+C\)