Integrate $\ln(2-x)dx$ I want to learn how to integrate this. If you could show me a step-by-step approach that would be awesome.
If you could also point me to some good tutorials on integration that would be icing on the cake.
What does $dx$ mean? I know its delta $x$ and very small strips or something (actually I'm not very sure...). If it is so minute ($\rightarrow 0$ ?), why isn't the whole thing just $0$ since $x * 0 = 0$ where $x$ is $\ln(2-x)$?
 A: Ok, since I got time, I'll bite, since you obviously lack the mathematical background to understand the rest of the answers. Definite integration is adding up infinitimestally small slices to add up to an area under a curve C. The definition of a definite integral is ;
Given a function  that is continuous on the interval $[a,b]$ we divide the interval into $n$ subintervals of equal width $\Delta x$, and from each interval choose a point $x_i$. Then $$\int_a^b f(x) \mathrm{d}x = \lim_{n \to \infty} \sum_{i=1}^n f(x_i) \Delta x$$
However, the integral you have requested is indefinite, which equals to the antiderivative of $f$ which means that after we solve this, if we differentiate the result we'll get $\ln(2-x)$. There are many integration techniques that you'll have to learn to tackle every integral, however remember that some functions have no antiderivative in terms of elementary functions and some don't have any at all!
Let's start with u-substitution(which is what you need for this problem of yours). We'll try to manipulate the stuff under the integral sign(which are named the integrand) in order to collapse this integral to an easier one that's in the table of integrals(or simplify it enough to re-manipulate it, but that's a little more complex, let's stick to your simple problem).
First of all you have to learn this formula;
If you let $u=2-x$ then you need $du$ which will be obtained by differentiation $u$ wrt. to $x$
So lets evaluate $\frac{d}{dx} u= \frac{d}{dx} (2-x)=\frac{d}{dx} 2 -\frac{d}{dx} x=0-1=-1$
Now if a mathematician sees this he'll cringe but for simplicity's sake, let's treat $dx$ algebraically. So you have $\frac{du}{dx}=-1 \iff du=-dx$
Let's go back to the integral and insert these new values;
$$\int \ln(2-x) dx= \int \ln (u)(-du)=\int -\ln (u) du$$
Now we have to learn a new integral property that states $$\int -f(x) dx=-\int f(x) dx$$
Let's use that new knowledge and get
$$\int -\ln (u) du=-\int \ln (u) du$$
Maybe this is too much, so I won't be introducing the integration by parts technique to solve this integral, as it is in integral tables anyway and we'll just copy the result. An important part to note is that we have to add an arbitrary constant to the result as if we differentiate that it will go away(remember, we get the antiderivative) anyway(but it could have been there, don't ever forget it).
Finally we get this;
$$-\int \ln (u) du=-u (\ln(u)-1)+\mathrm{constant} \stackrel{u=2-x}{=}-(2-x)(\ln(2-x)-1)+\mathrm{constant}$$
It's all algebra from here.
Now this all may seem a little bit too off or even counter-intuitive, but you're still on your way to learn a new branch in mathematics, named calculus, which I find quite beautiful and intriguing. I once started where you are and now my hobby is to try and solve difficult integrals.
EDIT; Here's another faster way to deal with your problem if you don't have any calculus background, and we don't want to introduce a lot of new stuff.
Let $f(x)=2-e^x$ which is the inverse of $g(x)=\ln(2-x)$. Then $f(y)=2-e^y$.
We'll use the formula $$\int f^{-1}(x)\,dx= x f^{-1}(x)-F\circ f^{-1}(x)+C$$ where $F$ is the antiderivative of $f$.
Let's evaluate $F(y)$(using some basic integral properties)
$$\int 2-e^y dy=\int 2 dy-\int e^y dy= 2y-e^y+c$$
Now lets evaluate(using function and logarithmic properties)
$$F\circ f^{-1}(x)=2(\ln(2-x))-e^{\ln(2-x)}+c=2\ln(2-x)-2+x+c=x+2 (\ln(2-x)-1) +c$$
We have all the parts for the formula so we just directly substitute them;
$$\int \ln(2-x) dx=x(\ln(2-x))-(x+2 (\ln(2-x)-1)))+c=$$
Which after algebraic manipulation gives the same result as above.
This is a special case where this theorem might simplify things for your case. It usually makes things more difficult so don't use it on every integral. Stick to normal integration techniques.
A: Integration by parts is the key:
$$\begin{eqnarray*}\int \log(2-x)\,dx &=& x\log(2-x)+\int \frac{x\,dx}{2-x} = x\left(\log(2-x)-1\right)+2\int\frac{dx}{2-x}\\&=&(x-2)\log(x-2)-x.\end{eqnarray*}$$
A: The first thing you could do is a u-sub. Let $2-x=t$ then the integral is of the form $-lnt*dt$ Now this is a standard integral which is $tlnt-t$ You can then back sub to get $(2-x)ln(2-x)-(2-x)$ and you can combine that -2 with a new constant.
Now how do we arrive at that $tlnt-t$? That you can do by doing another u-sub. Let $lnt=v$ so that $t=e^v$ and your integral becomes $ve^v*dv$
This can be done with Integration By Parts.
