Let D be bounded by the following lines:
$x+y=c$, $x+y=d$, $y=kx$, $y=lx$ with $ 0<c<d$ and $0<k<l$ and I need to transform this region into a rectangle
What I have done so far, the rectangle would be $R=[c,d] \times [k,l]$
I need to transform $x=x(u,v)$ and $y=y(u,v)$
I have defined $u=x+y$ and $v=\frac{x}{y}$
Now $x,y$ need to be expressed by $u$ and $v$.
I have $y=\frac{u}{v+1}$ and $x=\frac{u \cdot v}{v+1}$
Now to the Jacobian, $det\left(\frac{\partial(x,y)}{\partial(u,v)}\right)$=$\frac{-u}{(v+1)^2}$
then for the integral I have $\int\int_{D} f(x,y) d(x,y)= \int \int_{R} f(x(u,v), y(u,v))| \text{det}\frac{\partial(x,y)}{\partial(u,v)}| d(u,v)$
so it would be $=\int_{k}^{l} \int_{c}^{d} f(x(u,v), y(u,v)) | \text{det}\frac{\partial(x,y)}{\partial(u,v)}| dv du = \int_{k}^{l} \int_{c}^{d} \frac{-u}{(v+1)^2} du dv$ for the answer I get $\frac{-{(l-k)}{(d^2-c^2)}}{2(l+1)(k+1)}$
Could someone please tell me whether this was right or completely wrong?
Yes looks alright, other than a couple of minor things. The sign of your result is minus as $(l - k \gt 0, d^2 - c^2 \gt 0)$.
While $J(u,v) = -\frac{u}{(v+1)^2}$, $|J(u,v)| = \frac{u}{(v+1)^2}$ so you should not have negative sign inside the integral.
You set up $\frac{x}{y} = v$ and so the bounds of $v$ should be $(\frac{1}{l} \leq v \leq \frac{1}{k})$. You could have instead set up $\frac{y}{x} = v$ and then the bound of $v$ would have been $k \leq v \leq l$.
So $\int_{1/l}^{l/k} \int_{c}^{d} \frac{u}{(v+1)^2} du dv = \frac{{(l-k)}{(d^2-c^2)}}{2(l+1)(k+1)}$