Solving a Cauchy problem, differential equation I have the following Cauchy problem
\begin{cases} y'(x) + \frac{1}{x^2-1}y(x) = \sqrt{x+1} \\ y(0) = 0 \end{cases} 
I proceed by finding $e^{A(x)}  $ where $A(x)$ is the primitive of $a(x)= \frac{1}{x^2-1}$ :
$$\int A(x)dx=\int \frac{1}{x^2-1}dx= \frac{1}{2} \log\Big(\frac{|x-1|}{|x+1|}\Big)+c  $$
then  I obtain : $$e^{A(x)}=e^{\frac{1}{2} \log\Big(\frac{|x-1|}{|x+1|}\Big)}=\Big(\frac{|x-1|}{|x+1|}\Big)^{\frac{1}{2}}=\sqrt{\frac{|x-1|}{|x+1|} }$$
I have attempted to solve it in this way:
$$ \sqrt{\frac{|x-1|}{|x+1|} }\cdot y'(x) + \frac{1}{x^2-1}\cdot \sqrt{\frac{|x-1|}{|x+1|} }y = \sqrt{x+1}\cdot\sqrt{\frac{|x-1|}{|x+1|} }$$
$$\sqrt{\frac{|x-1|}{|x+1|} }*y(x) =\int \sqrt{\frac{x+1}{|x+1|}}\cdot\sqrt{|x-1|}dx$$
$$y(x) =\Big(\sqrt{\frac{|x-1|}{|x+1|} }\Big)^{-1}\cdot\int \sqrt{\frac{x+1}{|x+1|}}\cdot\sqrt{|x-1|}dx$$
Is it correct doing this? From here I am not sure how to proceed. Thanks in advance for any help.

Computing $$\mu(x)=e^{\int\frac{1}{x^2-1}dx}=\frac{\sqrt{1-x}}{\sqrt{1+x}}$$ then you will get
$$\int\frac{d}{dx}\left(\frac{\sqrt{1-x}y(x)}{\sqrt{x+1}}\right)=\int\sqrt{1-x}dx$$
