Question

$\int 7^{7^{7^x}}{.7}^{7^x}{.7}^{x}dx$ is equal to

Answer 1

Let $I=\int 7^{7^{7^x}}{7}^{7^x}{7}^{x}dx$

Let $t=7^x\Rightarrow \log t=x\log 7$

$\frac{1}{t}\frac{dt}{dx}=\log 7\frac{dx}{dx}$

$\Rightarrow dx=\frac{dt}{t\log 7}$

$\Rightarrow I=\int 7^{7^t}{7}^{t}t\frac{dt}{t\log 7}$

$\Rightarrow I=\int 7^{7^t}{7}^t\frac{dt}{\log 7}$

$\Rightarrow I=\frac{1}{\log 7}\int 7^{7^t}{7}^{t}dt$

Let $z=7^t\Rightarrow \log z=t\log 7$

$\frac{1}{z}\frac{dz}{dx}=\log 7\frac{dt}{dx}$

$\Rightarrow dt=\frac{dz}{z\log 7}$

$\Rightarrow I=\frac{1}{\log 7}\int 7^{z}z\frac{dz}{z\log 7}$

$\Rightarrow I=\frac{1}{{\left(\log 7\right)}^2}\int 7^{z}dz$ since $\int a^{x}dx=\frac{a^x}{\log a}$

$\Rightarrow I=\frac{1}{{\left(\log 7\right)}^3}{7}^z+c$ where $c$ is the constant of integration

$\Rightarrow I=\frac{1}{{\left(\log 7\right)}^3}{7}^{7^t}+c$ where $z=7^t$

$\Rightarrow I=\frac{1}{{\left(\log 7\right)}^3}{7}^{7^{7^z}}+c$ where $t=7^z$

Hence the result.