$\int 7^{7^{7^{x}}} 7^{7^{x}} 7^{x} \mathrm{~d} x=$

Question

Aicharya · Accepted Answer

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 $\frac{1}{y} \frac{d y}{d x}=(\log 7) \frac{d}{d x}\left(7^{7^{x}}ight)$ Let $z=7^{7^{x}} \Rightarrow \log z=7^{x} \log 7$ Differentiating w.r.t. $\mathrm{x}$, we get $\begin{aligned} & \frac{1}{z} \frac{d z}{d x}=7^{x}(\log 7)^{2} \ 	herefore \quad & \frac{d}{d x}\left(7^{7^{x}}ight)=7^{7^{x}} 7^{x}(\log 7)^{2} \Rightarrow \frac{d y}{d x}=7^{7^{7}} 7^{7^{x}} 7^{x}(\log 7)^{3} \ 	herefore \quad I=& \int \frac{d}{d x}\left(7^{7^{7^{x}}}ight) 	imes \frac{1}{(\log 7)^{3}} d x=\frac{7^{7^{x}}}{(\log 7)^{3}}+C \end{aligned}$

$\int 7^{7^{7^{x}}} 7^{7^{x}} 7^{x} \mathrm{~d} x=$

Solution