Answer
$\frac{\pi }{4} + \ln \sqrt 2 - 1$
Work Step by Step
$$\eqalign{
& {\text{Let the curves:}} \cr
& y = \tan x{\text{ and }}y = {\tan ^2}x,{\text{ on the interval }}\left[ {0,\frac{\pi }{4}} \right] \cr
& \tan x \geqslant {\tan ^2}x{\text{ on the interval }}\left[ {0,\frac{\pi }{4}} \right] \cr
& {\text{Then}}{\text{, the area is given by}} \cr
& A = \int_0^{\pi /4} {\left( {\tan x - {{\tan }^2}x} \right)} dx \cr
& {\text{Use the pythagorean identity }}{\sec ^2}\theta - 1 = {\tan ^2}\theta \cr
& A = \int_0^{\pi /4} {\left( {\tan x + 1 - {{\sec }^2}x} \right)} dx \cr
& {\text{Integrating }} \cr
& A = \left[ { - \ln \left| {\cos x} \right| + x - \tan x} \right]_0^{\pi /4} \cr
& {\text{Evaluating}} \cr
& A = \left[ { - \ln \left| {\cos \frac{\pi }{4}} \right| + \frac{\pi }{4} - \tan \left( {\frac{\pi }{4}} \right)} \right] - \left[ { - \ln \left| {\cos 0} \right| + 0 - \tan \left( 0 \right)} \right] \cr
& {\text{Simplifying}} \cr
& A = \left[ { - \ln \left| {\frac{{\sqrt 2 }}{2}} \right| + \frac{\pi }{4} - 1} \right] - \left[ 0 \right] \cr
& A = \frac{\pi }{4} + \ln \sqrt 2 - 1 \cr} $$
