Answer
Length $\simeq 3.235$
Work Step by Step
We have
$r = f\left( \theta \right) = 1 + \theta $, ${\ \ }$ $f'\left( \theta \right) = 1$
Using Eq. (7) the length of the curve for $0 \le \theta \le \frac{\pi }{2}$ is
$s = \mathop \smallint \limits_0^{\pi /2} \sqrt {{{\left( {1 + \theta } \right)}^2} + 1} {\rm{d}}\theta $
Let $u = 1 + \theta $. So, $du = d\theta $. The integral becomes
$s = \mathop \smallint \limits_1^{1 + \pi /2} \sqrt {1 + {u^2}} {\rm{d}}u$
Let $u = \tan v$. So, $du = {\sec ^2}vdv$. The integral becomes
$s = \mathop \smallint \limits_{{{\tan }^{ - 1}}1}^{{{\tan }^{ - 1}}\left( {1 + \pi /2} \right)} {\sec ^2}v\sqrt {1 + {{\tan }^2}v} {\rm{d}}v$
Since ${\sec ^2}v - {\tan ^2}v = 1$, we get
$s = \mathop \smallint \limits_{{{\tan }^{ - 1}}1}^{{{\tan }^{ - 1}}\left( {1 + \pi /2} \right)} {\sec ^3}v{\rm{d}}v$
From Eq. 14 of Section 8.2 (page 402) we have
$\smallint {\sec ^m}x{\rm{d}}x = \frac{{\tan x{{\sec }^{m - 2}}x}}{{m - 1}} + \frac{{m - 2}}{{m - 1}}\smallint {\sec ^{m - 2}}{\rm{d}}x$
So,
$s = \frac{{\tan v\sec v}}{2}|_{{{\tan }^{ - 1}}1}^{{{\tan }^{ - 1}}\left( {1 + \pi /2} \right)} + \frac{1}{2}\mathop \smallint \limits_{{{\tan }^{ - 1}}1}^{{{\tan }^{ - 1}}\left( {1 + \pi /2} \right)} \sec v{\rm{d}}v$
From Eq. 13 of Section 8.2 (page 402) we have
$\smallint \sec x{\rm{d}}x = \ln \left| {\sec x + \tan x} \right| + C$
So,
$s = \frac{{\tan v\sec v}}{2}|_{{{\tan }^{ - 1}}1}^{{{\tan }^{ - 1}}\left( {1 + \pi /2} \right)} + \frac{1}{2}\ln \left| {\sec v + \tan v} \right||_{{{\tan }^{ - 1}}1}^{{{\tan }^{ - 1}}\left( {1 + \pi /2} \right)}$
$s \simeq 3.235$
