Just a small integration question Please don't solve it in whole, just help me out with the first step here. Why do I have $d/dx$ and $dt$ at the same time? How to reform the problem so it looks more familiar. 
The way the problem is structured I find a bit odd. Thanks 
$$\displaystyle \dfrac{d}{dx}\left(\int_2^x\dfrac {dt}{t+ln(t)}\right)$$
 A: This problem is really asking whether or not you understand the statement of the Fundamental Theorem of Calculus. Note that the Fundamental Theorem of Calculus is often phrased in two parts (and not all sources agree on which is "part 1" and which is "part 2").
Part 1 is often stated: Suppose $f(x)$ is an integrable function. Then
$$ \frac{d}{dx} \int_0^x f(t) dt = f(x).$$
Part 2 is often stated: Suppose $f(x)$ is an integrable function and that $F(x)$ is a function such that $F'(x) = f(x)$. Then
$$ \int_a^b f(x) dx = F(b) - F(a).$$

I will also comment on some confusion in the problem. The $dt$ within the integrand serves to indicate what variable is being integrated. In this case, it's the $t$ variable. But the integral is a function of $x$. That is, there is a function in your question that we might call $g(x)$, given by
$$ g(x) = \int_2^x \frac{dt}{t + \ln t}.$$
This function has the interpretation "give the area under the graph of the function $1/(t + \ln t)$ from $2$ to $x$". So the $x$ and the $t$ are playing different roles: $t$ is what is being integrated, while $x$ is the variable of the overall function.
It would be very confusing to write
$$ \frac{d}{d{\color{#dd1111} x}} \int_2^{\color{#dd1111}x} \frac{d{\color{#5555ff} x}}{{\color{#5555ff}x} + \ln {\color{#5555ff} x}},$$
as the red $\color{#dd1111}x$ and the blue $\color{#5555ff}x$ have different meanings.
If it is helpful, I wrote a short note on introductory calculus for my students a few years ago.
