Answer
$$ y(x)=\frac{\tan ^{-1} x-\pi/4}{\sqrt{1+x^{2}}}$$
Work Step by Step
Given$$y^{\prime}+\frac{x}{1+x^{2}} y=\frac{1}{\left(1+x^{2}\right)^{3 / 2}}, \quad y(1)=0$$
This is a linear equation with $p(x) =\dfrac{x}{1+x^{2}}\ \ q(x) =\dfrac{1}{\left(1+x^{2}\right)^{3 / 2}}$, so
\begin{align*}
\mu(x)&=e^{\int p(x)dx}\\
&=e^{\int \frac{x}{1+x^{2}} d x}\\
&=e^{\frac{1}{2}\ln (1+ x^2)}\\
&= (1+ x^2)^{1/2}
\end{align*}
Then
\begin{align*}
y\mu(x) &=\int \mu(x)q(x)dx\\
(1+ x^2)^{1/2} y &=\int (1+ x^2)^{1/2} \dfrac{1}{\left(1+x^{2}\right)^{3 / 2}}dx\\
&= \int \dfrac{1}{1+x^{2}}dx\\
&=\tan^{-1}x+C
\end{align*}
Then $$y(x)=\frac{\tan ^{-1} x+C}{\sqrt{1+x^{2}}}$$
Since $y(1)=0 $, then $C= -\pi /4$, and hence
$$ y(x)=\frac{\tan ^{-1} x-\pi/4}{\sqrt{1+x^{2}}}$$
