General Solution of Differential Equation with Substitution Find the general solution of xy' + y = ${\sqrt{x^2+y^2}}$
So, I've tried using u=$\frac{y}{x}$, and that gets me to a point where I have.
$\int \frac{1}{\sqrt{1+u^2}-2u}du$ = $\int\frac{1}{x}dx$
The right side I know how to solve, I just have no idea what to do with this integral on the left.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
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 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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$\ds{t \equiv \root{1 + u^{2}} - u \implies u = {1 - t^{2} \over 2t}}$:

\begin{align}
\int{\dd u \over \root{1 + u^{2}} - 2u} & =
\int{\dd t \over t} - {2 \over 3}\int{2t \over t^{2} - 1/3}\,\dd t =
\ln\pars{\verts{t}} - {2 \over 3}\ln\pars{\verts{t^{2} - {1 \over 3}}}
\\[5mm] & =
\ln\pars{\verts{\root{1 + u^{2}} - u}} -
{2 \over 3}\ln\pars{\verts{\pars{\root{1 + u^{2}} - u}^{2} - {1 \over 3}}}
\end{align}
$$
= \bbx{\ln\pars{\verts{\root{1 + u^{2}} - u}} -
{2 \over 3}\ln\pars{\verts{2u^{2} + {2 \over 3} -2u\root{1 + u^{2}}}}
+ 
\pars{~\mbox{a constant}~}}
$$
