What is the integral of ##e^(7x)##?

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

We're going to use .

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!

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