Answer
$\displaystyle \frac{\ln 2}{2}$
Work Step by Step
$A=\displaystyle \int_{a}^{b}f(x)dx$
$A=\displaystyle \int_{0}^{\pi/4}\tan xdx$
... (see: integrals of the six basic trig. functions, p.333)
$A=\displaystyle \left[-\ln|\cos x|\right]_{0}^{\pi/4}=-[\ln|\cos\frac{\pi}{4}|-\ln|\cos 0|]$
$=-(\displaystyle \ln\frac{\sqrt{2}}{2}-\ln 1)$
$=-\displaystyle \ln(\frac{2^{1/2}}{2})=-\ln 2^{-1/2}=-(\frac{-1}{2})\ln 2$
$=\displaystyle \frac{\ln 2}{2}$
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