Answer
$$
S=\left\{(x, y) | x \geqslant 1,0 \leqslant y \leqslant 1 /\left(x^{3}+x\right)\right\}
$$
The shaded area is the region of interest.
$$
\begin{split}
\text { Area } & = \ \int_{1}^{\infty} \frac{1}{x^{3}+x} d x \\
& =\frac{1}{2} \ln 2
\end{split}
$$
Work Step by Step
$$
S=\left\{(x, y) | x \geqslant 1,0 \leqslant y \leqslant 1 /\left(x^{3}+x\right)\right\}
$$
The shaded area is the region of interest.
$$
\begin{split}
\text { Area } & = \int_{1}^{\infty} \frac{1}{x^{3}+x} d x \\
& = \lim _{t \rightarrow \infty} \int_{1}^{t} \frac{1}{x\left(x^{2}+1\right)} d x
\end{split}
$$
Now, decompose $\frac{1}{x\left(x^{2}+1\right)}$ into its partial fractions as follows,
$$
\frac{1}{x^{3}+x}=\frac{1}{x\left(x^{2}+1\right)}=\frac{A} { x}+ \frac{Bx+C} { \left(x^{2}+1\right)}
$$
$\Rightarrow \quad 1=A(x^{2}+1)+x(Bx+C)= (A+B)x^{2}+Cx+A $
by comparison of the coefficients we get $A=1, C=0, B=-1.$
Thus
$$
\frac{1}{x^{3}+x}=\frac{1} { x}- \frac{x} { \left(x^{2}+1\right)}
$$
Now integrate the decomposed version.
$$
\begin{split}
\text { Area } & = \int_{1}^{\infty} \frac{1}{x^{3}+x} d x \\
& = \lim _{t \rightarrow \infty} \int_{1}^{t} \frac{1}{x\left(x^{2}+1\right)} d x \\
& = \lim _{t \rightarrow \infty} \int_{1}^{t}\left(\frac{1}{x}-\frac{x}{x^{2}+1}\right) d x \\
& = \lim _{t \rightarrow \infty} \left[\ln |x|-\frac{1}{2} \ln \left|x^{2}+1\right|\right]_{1}^{t} \\
& = \lim _{t \rightarrow \infty} \left[ \ln \frac{x}{\sqrt{x^{2}+1}}\right]_{1}^{t}\\
& = \lim _{t \rightarrow \infty} \left(\ln \frac{t}{\sqrt{t^{2}+1}}-\ln \frac{1}{\sqrt{2}}\right) \\
& = \ln 1-\ln 2^{-1 / 2}\\
&=\frac{1}{2} \ln 2
\end{split}
$$
