Burdett and Mortensen (1998), Equation (22), Integral by parts question Suppose $F(x)$ and $H(x)$ are both cumulative prob dist functions on the support of $[b_0,b_1]$ and we know followings:

$$u(x|F)=\int_{b_0}^{x}\frac{m}{1+k[1-F(b)]} dH(b)$$ 

and from this 

$$[1+k(1-F(x))]du(x|F)=mdH(x)$$ 

We also know that 

$$G(w)(m-u(b_1|F))=\frac{k\int_{b_0}^w [F(w)-F(x)]du(x|F)}{1+k(1-F(w))}$$

Then in the paper, authors get following which is equation [22] in the paper

$$\frac{(m-u(b_1|F))dG(w)}{dF(w)}=\frac{kmH(w)}{[1+k[1-F(w)]]^2}$$

I really don't understand how the authors got this. Can somebody help me?
The authors apply in the second step the exact same rules they used in the first step, regarding the calculation of the total differential of an integral, taking also into account that the integral is taken over a function of the integrating variable and not just the integrating variable itself.  

The first step is 

$$d[u(x|F)]=d\left[\int_{b_0}^{x}\frac{m}{1+k[1-F(b)]} dH(b)\right]$$

$$=\frac{m}{1+k[1-F(x)]}dH(x) \tag{1}$$

and re-arranging we get the first result. Applying the same logic, we have

$$(m-u(b_1|F))\cdot d\big[G(w)\big]=kd\left[\frac{\int_{b_0}^w [F(w)-F(x)]du(x|F)}{1+k(1-F(w))}\right] $$

We are considering the differential with respect to $w$, since this is the argument of the $G$ function. To clearly indicate this we will write $d_w$. To clarify the picture write

$$kd_w\left[\frac{\int_{b_0}^w [F(w)-F(x)]du(x|F)}{1+k(1-F(w))}\right] \equiv k d_w\left(\frac{A(w)}{B(w)}\right) = k\frac{B(w)d_wA(w) - A(w)d_wB(w)}{[B(w)]^2}$$

We see that the denominator is exactly as the final expression of the authors, so it is of no concern. We turn to the numerator.
One by one,

$$ d_wA(w) = d_w\left[\int_{b_0}^w [F(w)-F(x)]du(x|F)\right] $$

$$= [F(w)-F(w)]du(w|F) + \left[\int_{b_0}^w d_w[F(w)-F(x)]du(x|F)\right]$$

$$= 0 + d_wF(w) \int_{b_0}^w du(x|F) + 0$$

We applied here Leibniz rule for differentiation w.r.t. a limit of integration under the integral when the integrand is also a function of the integrating limit. The second zero result comes from the fact that $d_wF(x) = 0$ ($F(x)$ does not include $w$, so as we change $w$ the change in $F(x)$ is zero). So 

$$B(w)d_wA(w) = [1+k(1-F(w)]d_wF(w) \int_{b_0}^w du(x|F) $$

$$= d_wF(w)\int_{b_0}^w [1+k(1-F(w))]du(x|F) \tag{2}$$

where we have moved the term inside the integral since it does not depend on the integrating variable which is $x$.

Going to the second term, we have

$$d_wB(w) = -kd_wF(w)$$

So

$$-A(w)d_wB(w) = d_wF(w) \int_{b_0}^w k[F(w)-F(x)]du(x|F) \tag{3}$$

Therefore

$$B(w)d_wA(w) -A(w)d_wB(w) = d_wF(w)\int_{b_0}^w [1+k(1-F(w)]du(x|F) \\+ d_wF(w) \int_{b_0}^w k[F(w)-F(x)]du(x|F)$$

The $d_wF(w)$ is a common factor and the two integrals can be written as one due to linearity, so 

$$B(w)d_wA(w) -A(w)d_wB(w)=\\ = d_wF(w)\int_{b_0}^w\Big([1+k(1-F(w)] + k[F(w)-F(x)]\Big)du(x|F)$$

The term $kF(w)$ in the integrand cancels off and we are left with 

$$B(w)d_wA(w) -A(w)d_wB(w) = d_wF(w)\int_{b_0}^w[1+k(1-F(x))]du(x|F)$$

But using the re-arranged $(1)$ and substituting we get

$$B(w)d_wA(w) -A(w)d_wB(w) = d_wF(w)\int_{b_0}^wmdH(x)$$

$$= md_wF(w)\cdot \Big(H(x) \big|_{b_0}^w\Big) = md_wF(w)\cdot \Big(H(w) - H(b_0)\Big)$$

But since $H()$ is a CDF in $(b_0, b_1)$, it follows that $H(b_0) =0$.

And we are done. In short, don't be afraid to push through the mathematical calculations.