# Rewriting a function so it is in a form that can be easily integrated.

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 $x^n$.

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One of the other rules for indefinite integrals is that the integral of $a \cdot g(x)$ for some constant $a$ is $a \cdot G(x)$, where $G(x)$ is the indefinite integral of $g(x)$. So instead of calculating the integral of $\frac{2}{3} x$, try finding the integral for $x$ and then multiplying by $\frac{2}{3}$. – TMM Mar 3 '12 at 15:49
Excellent, thank you very much. If you put that as an answer i'll accept it. I'll get my head around all of these rules one day. – Aesir Mar 3 '12 at 16:03

Writing $I(f(x))$ for the indefinite integral of $f(x)$, we need the following rules:
\begin{align}I(f(x) + g(x)) = I(f(x)) + I(g(x)) \tag{1}\end{align} $$I(a \cdot f(x)) = a \cdot I(f(x)), \quad \text{for constants a} \tag{2}$$ $$I(x^n) = \frac{1}{n+1}{x^{n+1}} + c \tag{3}$$
Applying these to your function $f(x)$, we get:
$$\begin{eqnarray}I(f(x)) &=& I\left(\frac{2}{3} x + 3\right) \\ &\stackrel{(1)}{=}& I\left(\frac{2}{3} x\right) + I(3) \\ &\stackrel{(2)}{=}& \frac{2}{3} \cdot I(x) + 3 \cdot I(1) \\ &\stackrel{(3)}{=}& \frac{2}{3} \cdot \left(\frac{1}{2} x^2 + c_1\right) + 3 \cdot (x + c_2) \\ &=& \frac{1}{3} x^2 + 3x + \left(\frac{2}{3} c_1 + 3 c_2\right) \\ &=& \frac{1}{3} x^2 + 3x + c, \end{eqnarray}$$
where we wrote $c = \frac{2}{3} c_1 + 3 c_2$ for a new constant.