Answer
$$L = \sqrt 2 \ln \left( {\sqrt 2 + 1} \right)$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {t,\ln \sec t,\ln \left( {\sec t + \tan t} \right)} \right\rangle ,\,\,\,\,\,for\,\,\,\,\,0 \leqslant t \leqslant \frac{\pi }{4} \cr
& {\text{Calculate }}{\bf{r}}'\left( t \right){\text{ and }}\left| {{\bf{r}}'\left( t \right)} \right| \cr
& {\bf{r}}'\left( t \right) = \left\langle {t,\ln \sec t,\ln \left( {\sec t + \tan t} \right)} \right\rangle \cr
& {\bf{r}}'\left( t \right) = \left\langle {1,\tan t,\sec t} \right\rangle \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt {{{\left( 1 \right)}^2} + {{\left( {\tan t} \right)}^2} + {{\left( {\sec t} \right)}^2}} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt {1 + {{\tan }^2}t + {{\sec }^2}t} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt {{{\sec }^2}t + {{\sec }^2}t} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt 2 \sec t \cr
& {\text{The Arc Length is }}\int_a^b {\left| {{\bf{r}}'\left( t \right)} \right|} dt \cr
& L = \int_0^{\pi /4} {\sqrt 2 \sec t} dt \cr
& {\text{Integrate}} \cr
& L = \sqrt 2 \left[ {\ln \left| {\sec t + \tan t} \right|} \right]_0^{\pi /4} \cr
& L = \sqrt 2 \left[ {\ln \left| {\sec \left( {\frac{\pi }{4}} \right) + \tan \left( {\frac{\pi }{4}} \right)} \right| - \ln \left| {\sec \left( 0 \right) + \tan \left( 0 \right)} \right|} \right] \cr
& L = \sqrt 2 \left[ {\ln \left( {\sqrt 2 + 1} \right) - \ln \left( {1 + 0} \right)} \right] \cr
& L = \sqrt 2 \ln \left( {\sqrt 2 + 1} \right) \cr} $$
