Q. If $x$ is any arbitrary constant, then $\int_{}^{}{2^{2^{2^{x}}}.2^{2^{x}}.2^{x}}dx$ is equal to
A) $\frac{\int_{}^{}{2^{2}}^{x}dx}{\left( \log 2 \right)^{3}} + c$
B) $\frac{\int_{}^{}{2^{2^{2}}}^{x}dx}{\left( \log 2 \right)^{3}} + c$
C) $\int_{}^{}{2^{2^{2}}}^{x}\left( \log 2 \right)^{3}dx + c$
D) None of these
Explanation
(d)
Let $I = \int_{}^{}{2^{2^{2^{x}}}.2^{2^{x}}.2^{x}}dx$
Put $2^{2^{2^{x}}} = t\ \Rightarrow \ 2^{2^{2^{x}}}.2^{2^{x}}.2^{x}\left( \log 2 \right)^{3}dx = dt$
$\therefore\ I = \int_{}^{}\frac{dt}{\left( \log 2 \right)^{3}} = \frac{t}{\left( \log 2 \right)^{3}} = \frac{2^{2^{2^{x}}}}{\left( \log 2 \right)^{3}} + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.