e constant integration The integration of a constant is the constant itself plus some other "constant"
$\int{e^xdx}=e^x+C$
So, the $\int{e^{5x}dx}$ must be:
$$e^{5x}dx=e^{5x}+C$$
This is true if e here is a constant, generally expressing Euler constant: 2.7182...
Am I right?
I run the integral with Wolfram Alpha, this is the result:
$$\frac{e^{5x}}{5}+C$$
Is this happen because Wolfram Alpha do not understand e as a constant?
 A: "The integration of a constant is the constant itself plus some other "constant" "
Is this true?
$\int 4 dx=4x\neq 4+C$
Therefore, it is not true.
Also $\frac d{dx} e^x$ can be thought of like this: Take the derivative of $e^x$, which is $e^x$, and then using the chain rule, take the derivative of $x$, which would give you $e^x (1)=e^x$.
In your example $\frac d{dx} e^{5x}=e^{5x}(5)=5e^{5x}$ by similar logic.
Also note that $e^x$ is not a constant. This is because depending on the value of $x$, its value changes. In a constant, no matter what $x$ value I sub in, it should remain the same. (Example: Given $f(x)=3$, then $f(1)=f(56)=f(9999)=f(0.45)=f(-9)=3)$. Try graphing it; a constant will have a graph that is a horizontal line.
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Edit: Fixing antiderivative
A: No this happens because the primitive of $e^{5x}$ is $\frac{e^{5x}}{5}$. This comes from the general formula $(f\circ g)'=g'*(f'\circ g)$.
A: $\int e^{ax}dx = \frac{1}{a} e^{ax} + C$ . Mathematica is correct
A: Remember that:
$$\int e^{nx}\ dx=n^{-1}e^{nx}+C$$
Remember the chain rule?
