Evaluate the integral: $\int_{0}^3\frac{dx}{5x+1}$ $\int_{0}^3\frac{dx}{5x+1}$
So I cannot evaluate this integral because we do not have any rules for taking the anti-derivative of a quotient of functions. Thus, we need to separate this into two separate fractions before we can take the anti-derivative. 
How do we do this? 
$\frac{1}{5x+1}$ I can't transfer the binomial to the top unless I can do the following, and if so please make me aware that I am right. 
$\frac{(5x+1)^{-1}}{1}$
Is this "illegal math"
 A: A minor point: It is unnecessary to write $\frac{(5x+1)^{-1}}{1}$ with a $1$ in the denominator. For any number $a$ it is always true that $a = \frac{a}{1}$ so we write simply $(5x+1)^{-1}$. 
As for your main question, you probably know by know that you should make the substitution $5x+1 = t$. You are correct that there is no specific rule to integrate the quantity that you have. So what you want to do is pattern-match what you have to an integral form that you recognize. Ideally you recall that the derivative of $\ln(x)$ is $\frac{1}{x}$, and in that capacity you recognize the integral $\int \frac{1}{x} \text{d}x$. This is why the substitution $5x+1 = t$ is useful, because it converts your problem into one that you can easily anti-differentiate. Plugging in this substitution yields $$\int_0^3 \frac{1}{t} \text{d}x$$ but before continuing there are still two things to address. $(1)$ We need $\text{d}t$ not $\text{d}x$, as we are now trying to integrate with respect to $t$, and $(2)$ the integral limits from $0$ to $3$ were meant for your original quantity, not your substituted quantity. These will need to change to match your $t$ substitution.
Getting $\text{d}t$ is pretty easy. Take $5x+1 = t$ and differentiate both sides. You'll get $$5\text{d}x = 1\text{d}t \implies \text{d}x = \frac{1}{5}\text{d}t$$
For the new limits again take $5x+1 = t$. When $x = 0$ we have $5(0) +1 = t \implies t = 1$ and when $x = 3$ we have $5(3)+1 = t \implies t = 16$, so your new limits of integration are from $1$ to $16$. We now have everything we need to set up the new integral. $$\begin{align} \int_0^3 \frac{1}{5x+1}\text{d}x &\equiv \int_1^{16} \frac{1}{t}\left(\frac{1}{5}\text{d}t\right) \\ &= \frac{1}{5} \int_1^{16} \frac{1}{t}\text{d}t \end{align}$$
If this process is new to you, take note. It's (probably) the most common method you will use to solve integrals: Pattern-match your integral to one that you know how to anti-differentiate, make a substitution, solve for your infinitesimal in terms of the substitution, solve for the new limits of integration and lastly integrate.
A: Notice, let $5x+1=u\implies 5dx=du$ $$dx=\frac{du}{5}$$
For lower limit $x=0\implies u=5(0)+1=1$
For upper limit $x=0\implies u=5(3)+1=16$
Now, substituting the corresponding values, we get
$$\int_{0}^{3}\frac{dx}{5x+1}=\int_{1}^{16}\frac{1}{u}\frac{du}{5}$$
$$=\frac{1}{5}\int_{1}^{16}\frac{du}{u}$$ $$=\frac{1}{5}[\ln|u|]_{1}^{16}$$
$$=\frac{1}{5}[\ln|16|-\ln|1|]=\frac{1}{5}[\ln|16|-0]$$ $$=\color{}{\frac{1}{5}\ln(16)}=\color{}{\frac{1}{5}\ln(2^4)}$$ $$=\color{red}{\frac{4}{5}\ln 2}$$
A: Hint:
with the substitution $5x+1=t$ you have $5dx=dt$ and the integral becomes:
$$
\int \dfrac{1}{5}\cdot \dfrac{dt}{t}
$$
can you integrate?
A: Let $5x+1 = t\;,$ Then $d(5x+1) = dt\Rightarrow \displaystyle \frac{d}{dx}(5x+1)dx = dt$
So we get $$\displaystyle 5dx=dt\Rightarrow dx = \frac{1}{5}dt$$ and changing limit
We get $$\displaystyle \frac{1}{5}\int_{0}^{16}\frac{1}{t}dt = \frac{1}{5}\left[\ln |t|\right]_{1}^{16} = \frac{1}{5}\left[\ln|16|-\ln|1|\right] = \frac{1}{5}\ln(16)$$
A: No it isn't illegal to do that, it is the same thing.  To solve sub $u=5x+1$ then $du=5dx$ which means $1/5du=dx$  Your integral becomes $\frac 1 5\int_{1}^{16}\frac{du}{u}$ and use the fact that $\int\frac {du} u=ln x$, should be absolute value of x in that last expression.
