Find the area bounded by the arc of the curve y = arctan

Find the area bounded by the arc of the curve $y = \arctan\left(1 + \frac{1}{x}\right)$ and the coordinate axes.

We will check where the curve meets the axes and examine whether the resulting region is actually bounded.

Step 1: Determine the domain and key limits

The curve $y=\arctan\left(1+\frac1x\right)$ is defined for $x\neq0$. As $x\to+\infty$: $1+\frac1x\to1$, so $y\to\arctan1=\frac\pi4$. As $x\to0^+$: $1+\frac1x\to+\infty$, so $y\to\frac\pi2$. As $x\to0^-$: $1+\frac1x\to-\infty$, so $y\to-\frac\pi2$.

Step 2: Find where the curve meets the axes

Intersection with $y=0$:

\[\arctan\left(1+\frac1x\right)=0 \implies 1+\frac1x=0 \implies x=-1\]

So the curve crosses the $x$-axis only at $x=-1$; it never crosses $y=0$ for $x>0$ (there it stays strictly between $\frac\pi4$ and $\frac\pi2$). The curve is undefined at $x=0$, so it never meets the $y$-axis either.

Step 3: Recognize the region is unbounded

For $x>0$, the curve stays entirely above the $x$-axis and approaches the horizontal asymptote $y=\frac\pi4$ as $x\to\infty$ — it never comes back down to close off a finite region with the axes. The natural area to test for finiteness is then:

\[A = \int_0^{\infty} \arctan\left(1+\frac1x\right) dx\]

Step 4: Show the integral diverges

Since $\arctan\left(1+\frac1x\right)\to\frac\pi4$ as $x\to\infty$, for any large $x$ the integrand is bounded below by a constant close to $\frac\pi4>0$. An integral of a function that tends to a positive constant over an infinite interval necessarily diverges:

\[\int_0^{\infty} \arctan\left(1+\frac1x\right) dx = \infty\]

Formally, integrating by parts with $u=\arctan\left(1+\frac1x\right)$, $dv=dx$ gives a boundary term $\left[x\arctan\left(1+\frac1x\right)\right]_0^a\to a\cdot\frac\pi4$ as $a\to\infty$, which itself diverges.

Conclusion

\[\boxed{A = \infty \text{ — the region bounded by the curve and the coordinate axes for } x>0 \text{ has infinite area}}\]

A bounded variant

If instead a finite right-hand boundary is imposed, say the strip $x\in[0,1]$, the resulting proper integral $\int_0^1\arctan\left(1+\frac1x\right)dx$ is finite; numerically it evaluates to approximately $1.238$ square units, though it has no elementary closed form.

Summary

For the unbounded region with $x\in(0,\infty)$:

\[\boxed{A = \infty}\]

Key insight: A curve that approaches a nonzero horizontal asymptote can never bound a finite area together with the $x$-axis alone over an infinite strip — the "coordinate axes" here do not actually close off the region for $x>0$, since the curve stays strictly above $y=0$.