Find the general integral of the equation…
Find the general integral of the equation $(y'+1)\ln\dfrac{y+x}{x+3}=\dfrac{y+x}{x+3}$.
Explanation of Method
Both sides involve the same ratio, so substitute $u=\dfrac{y+x}{x+3}$; differentiating this relation expresses $y'+1$ in terms of $u$ and $u'$, collapsing the equation to a separable one.
Step 1: Substitute $u=\dfrac{y+x}{x+3}$
From $y=u(x+3)-x$, $y'+1=(x+3)u'+u$. Substituting into the equation:
\[(x+3)u'\ln u=u-u\ln u=u(1-\ln u).\]
Step 2: Separate and integrate
With $v=\ln u$,
\[\frac{\ln u}{u(1-\ln u)}\,du=\frac{dx}{x+3} \quad\Longrightarrow\quad \int\frac{v}{1-v}\,dv=\ln|x+3|+C.\]
Since $\dfrac{v}{1-v}=-1+\dfrac{1}{1-v}$, the left side integrates to $-v-\ln|1-v|=-\ln u-\ln|1-\ln u|$, giving
\[\ln[u(1-\ln u)]=-\ln|x+3|+K \quad\Longrightarrow\quad (x+3)\,u(1-\ln u)=C.\]
Step 3: Back-substitute $u=\dfrac{y+x}{x+3}$
Since $(x+3)u=y+x$,
\[\boxed{\,(y+x)\!\left[1-\ln\frac{y+x}{x+3}\right]=C.}\]
Result
Final answer:
\[\boxed{\,(y+x)\left[1-\ln\dfrac{y+x}{x+3}\right]=C.}\]