I am trying to find an integral of the function $f(x)=\frac{2}{3}x+3$. I am a bit stumped on how to find the integral of $\frac{2}{3}x$, specifically.
I looked at the answer thinking I could work backwards but got stuck doing this as well. I suspect I am missing something pretty obvious.
The answer provided is: $\frac{1}{3}x^2+3x+c$
So by using the rule for indefinite integrals I have been given for $x^n$:
$\frac{1}{n+1}x^{n+1}+c$
I thought I could just take the answer and work out $n$ by reversing this formula. But if my answer is $\frac{1}{3}x^2+c$ then by the rule above $n=1$ and $n=2$ for the answer provided.
This can't be right, which means I am using the rule incorrectly. I think I have to rewrite $\frac{2}{3}x$ as $x$ to the power of something say $y$ so they are equivalent i.e. $\frac{2}{3}x=x^y$ but I do not understand how to do this.
I am basing this on the answer provided being in the form of the rule stated above which applies to functions in the form of $x^n$.