0
$\begingroup$

I have to solve $\displaystyle \frac{dy}{dx}=\left(\frac{x+y+1}{x+y+3}\right)^{2}$.

i am taking $x+y=u.$ So i get $\displaystyle \frac{du}{dx}=\left(\frac{u+1}{u+3}\right)^{2}+1$. After this i dont know how to integrate this.

1 Answers 1