Differential equation : $y' = (x+1)/(xy+x)$ So, I have the following differential equation to solve : $$y' = \frac{x+1}{xy+x}$$
After several steps, I get here :

$t^2 + 2t = 2x + 2ln(x) + c$

How do I isolate $t ?$

thank you!

By the way, $ t=f(x)=y$
Start by dividing through by $x$:
$$
y'=\frac{1+\frac{1}{x}}{y+1} \\
y'(y+1)=1+\frac{1}{x} \\
\int y'(y+1)dx=\int 1+\frac{1}{x}dx \\
\frac{y^2}{2}+2y=2x+2\ln(x)+c~\text{for constant $c$} \\
y^2+2y+1=2x+2\ln(x)+2c+1 \\
(y+1)^2=2x+2\ln(x)+2c+1 \\
y+1=\pm \sqrt{2x+2\ln(x)+2c+1} \\
y=\pm \sqrt{2x+2\ln(x)+2c+1}-1.~_{\square}
$$
let $W$ denote Lambert's W-function. after taking exponentials of both sides:
$$
Ae^{\frac12(t+1)^2} = xe^x =W^{-1}(x)
$$
so we have
$$
x = W(Ae^{\frac12(t+1)^2})
$$
if this is correct, then we have:
$$
t = \sqrt(2\log( A^{-1}W^{-1}(x)))-1
$$