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dy/y = x/(x^2+a^2) dx
integral each side...

\(\int\frac{dy}{y}=\int\frac{x\,dx}{a^2+x^2}\)
\(\ln|y|=\frac{1}{2}\ln(a^2+x^2)+C\)

oo ok..and we just solve for y?

use e

\(|y|=Ae^{\ln(\sqrt{a^2+x^2})}=A\sqrt{a^2+x^2}\)
\(y=\pm A\sqrt{a^2+x^2}=A\sqrt{a^2+x^2}\)

what is the capital A??

That is to replace \(e^C\) which is just another constant.

oo for e^(1/2)..?

no
for e^c

\(\frac{1}{2}\ln (a^2+x^2)=\ln(a^2+x^2)^{1/2}\) by the property of \(\ln\).

yes..i understand it now thanks guys!

I think we need to add small detail that \(A >0\).