int 7^{7^{7^{x}}} 7^{7^{x}} 7^{x} ,d x= | vedclass

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(A) Let $I = \int 7^{7^{7^{x}}} 7^{7^{x}} 7^{x} dx$.Let $u = 7^{x}$. Then $du = 7^{x} \log 7 dx$,so $7^{x} dx = \frac{du}{\log 7}$.The integral become

Vedclass · Accepted Answer

$\frac{7^{7^{7^{x}}}}{(\log 7)^{3}}+C$

Vedclass · Answer

(A) Let $I = \int 7^{7^{7^{x}}} 7^{7^{x}} 7^{x} dx$.Let $u = 7^{x}$. Then $du = 7^{x} \log 7 dx$,so $7^{x} dx = \frac{du}{\log 7}$.The integral becomes $I = \int 7^{7^{u}} 7^{u} \frac{du}{\log 7} = \frac{1}{\log 7} \int 7^{7^{u}} 7^{u} du$.Let $v = 7^{u}$. Then $dv = 7^{u} \log 7 du$,so $7^{u} du = \frac{dv}{\log 7}$.Substituting this into the integral,we get $I = \frac{1}{\log 7} \int 7^{v} \frac{dv}{\log 7} = \frac{1}{(\log 7)^{2}} \int 7^{v} dv$.Since $\int 7^{v} dv = \frac{7^{v}}{\log 7} + C$,we have $I = \frac{7^{v}}{(\log 7)^{3}} + C$.Substituting back $v = 7^{u} = 7^{7^{x}}$,we get $I = \frac{7^{7^{7^{x}}}}{(\log 7)^{3}} + C$.

$\int 7^{7^{7^{x}}} 7^{7^{x}} 7^{x} \,d x=$

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