$\int 7^{7^{7^x}} . 7^{7^x} . 7^x d x$ is equal to
Answer & explanation
Correct answer: option 1
$I=\int 7^{7^{7^x}} . 7^{7^x} . 7^x d x$
let $y=7^x$ so $dy=7^x\log 7dx$
$⇒I=\int\frac{7^{7^y}7^y dy}{\log 7}$
let $z=7^y$ so $dz=7^y\log 7dy$
$⇒I=\int\frac{7^zdz}{(\log 7)^2}=\frac{7^z}{(\log 7)^3}+C=\frac{7^{7^{7^x}}}{\left(\log 7\right)^3}+C$