Differential Equation. $\frac{dy}{dx}=(x+y-1)+\frac{x+y}{log(x+y)}$ $\frac{dy}{dx}=(x+y-1)+\frac{x+y}{log(x+y)}$  
The question is from IIT entrance exam practice material. I have tried substituting  (x+y=t) but was stuck after some process.  Please help me out with this. 
The solution is : $(1+log(x+y)-log{(1+log(x+y))=x+C}$
 A: Set $v=x+y$. We then have
$$\dfrac{dv}{dx} = 1 + \dfrac{dy}{dx} = 1 + (v-1) + \dfrac{v}{\ln(v)} = v + \dfrac{v}{\ln(v)}$$
Hence,
$$\dfrac{\ln(v) dv}{v(1+\ln(v))} = dx$$
Setting $v = e^t$, we get
$$\dfrac{tdt}{1+t} = dx$$
You should be able to finish it off from here.
A: Let $t = x + y \Rightarrow \frac{dt}{dx} = 1 + \frac{dy}{dx} $. The equation becomes
$$ \frac{dt}{dx} = t + \frac{t}{\ln t} = \frac{t(\ln t + 1)}{\ln t} $$ 
Using seperation of variables
$$ \frac{\ln t}{t(\ln t + 1)} dt = dx $$
$$ \int \frac{\ln t}{t(\ln t + 1)} dt = x + C $$
To solve the left integral, let $u = \ln t \Rightarrow du = \frac{dt}{t}$
$$ \int \frac{u}{u + 1}du = x + C $$
$$ u - \ln (u + 1) = x + C$$
$$ \ln (x + y) - \ln \big(\ln(x+y) + 1 \big) = x + C $$
A: I substituted $x+y=t$ than $dy/dx=dt/dx-1$
$dt/dx=t+t/lnt$
$dt/dx=t(lnt+1)/lnt$
$\int \frac{lnt}{t(lnt+1)} dt = x$
than I substitute  
$lnt=u$ 
$du/dt=1/t$
which gives me 
$\int\frac{u}{u+1}du$=$x$
hence,   
$x+c=u-ln(u+1)$
solution: $x+c=ln(x+y)-ln(ln(x+y)+1)$
I don't know where I missed the 1 on RHS as the answer is 
$(1+ln(x+y)-ln{(1+ln(x+y))=x+C}$
