$\int 7^{2 x} d x=\varphi(x)+c$ then $\varphi(x)=$

The integral to be evaluated is $I = \int 7^{2x} dx$.

Using the standard integration formula $\int a^{u} du = \dfrac{a^{u}}{\ln a} + c$, where $a = 7$ and $u = 2x$.

Since $u = 2x$, $du = 2 dx$, which implies $dx = \dfrac{du}{2}$.

Substituting these into the integral:

$I = \int 7^{u} \dfrac{du}{2} = \dfrac{1}{2} \int 7^{u} du$.

$I = \dfrac{1}{2} \left( \dfrac{7^{u}}{\ln 7} \right) + c$.

Substituting $u = 2x$ back into the expression:

$I = \dfrac{7^{2x}}{2 \ln 7} + c$.

Since $\ln 7 = \log_{e} 7$, which is often denoted as $\log 7$ in this context, we have $\varphi(x) = \dfrac{7^{2x}}{2 \log 7}$.

Answer: $\dfrac{7^{2 x}}{2 \log 7}$