How to add two dependent random variables? I know, I can't use convolution. 
I have two random variables A and B and they're dependent.
I need Distributive function of A+B
 A: As vinux points out, one needs the joint distribution of $A$ and $B$, and 
it is not obvious from OP Mesko's response "I know Distributive function of A and B"
that he is saying he knows the joint distribution of A and B: he may well
be saying that he knows the marginal distributions of A and B.  However,
assuming that Mesko does know the joint distribution, the answer is given below.
From the convolution integral in OP Mesko's comment (which is wrong, by the way), it could be inferred that 
Mesko is interested in jointly continuous random variables $A$ and $B$  with joint probability density function $f_{A,B}(a,b)$. In this case,
$$f_{A+B}(z) = \int_{-\infty}^{\infty} f_{A,B}(a,z-a) \mathrm da
= \int_{-\infty}^{\infty} f_{A,B}(z-b,b) \mathrm db.$$
When $A$ and $B$ are independent, the joint density function factors into the
product of the marginal density functions: $f_{A,B}(a,z-a)=f_{A}(a)f_{B}(z-a)$ 
and we get the more familiar
convolution formula for independent random variables.  A similar result
applies for discrete random variables as well.
Things are more complicated if $A$ and $B$ are not jointly continuous, or
if one random variable is continuous and the other is discrete.  However,
in all cases, one can always find the cumulative probability distribution
function $F_{A+B}(z)$ of $A+B$ as the total probability mass in the region of 
the plane specified as $\{(a,b) \colon a+b \leq z\}$ and compute the probability
density function, or the probability mass function, or whatever, from the
distribution function.  Indeed the above formula is obtained by writing
$F_{A+B}(z)$ as a double integral of the joint density function over the
specified region and then "differentiating under the integral
sign.''
A: Beforehand , I don't know if what i'm saying is correct but I got stuck on the same problem and I tried to solve it in this way:
express the joint distribution using the heaviside step function:
$$
f_{A,B}(a,b)=(a+b) H(a,b) H(-a+1,-b+1) 
$$
or equivalently
$$
f_{A,B}(a,b)=(a+b)(H(a)-H(a-1))(H(b)-H(b-1))
$$
Now you can perform the integral without caring about limits of integration.
This is the wolfram rapresentation of the joint : A
Computing the integral I have : B
Plotted : C
That's the function :
$$ f(z)=\left\{\begin{matrix}
z^2 \qquad \qquad  \; \quad for \quad 0\leq z \leq1\\ 1-(z-1)^2 \quad for  \quad 1\leq z \leq 2 \\ 0 \qquad \quad \; otherwise
\end{matrix}\right. $$ 
and it's normalized as you can easily check.
