It’s #1/7e^(7x)#
What you want to calculate is:
#int e^(7x)dx#

We’re going to use

u-substitution

.

Let #u = 7x#
Differentiate (derivative) both parts:
#du = 7dx#
#(du)/7 = dx#
Now we can replace everything in the integral:
#int 1/7 e^u du#
Bring the constant upfront
#1/7 int e^u du#
The integral of #e^u# is simply #e^u#
#1/7e^u#
And replace the #u# back
#1/7e^(7x)#

There’s also a shortcut you can use:
Whenever you have a function of which you know the integral #f(x)#, but it has a different argument
#=># the function is in the form #f(ax+b)#
If you want to integrate this, it is always equal to #1/a*F(ax+b)#, where #F# is the integral of the regular #f(x)# function.

In this case:
#f(x) = e^x#
#F(x) = int e^x dx = e^x#
#a = 7#
#b = 0#
#f(ax+b) = e^(7x)#
=> # int e^(7x)dx = 1/a*F(ax+b) = 1/7*e^(7x) #

If you use it more often, you will be able to do all these steps in your head.
Good luck!