79.计算定积分int_(0)^1(ln(1+x))/(1+x^2)dx.

令 $ x = \tan t $，则 $ dx = \sec^2 t \, dt $，积分变为
$I = \int_{0}^{\frac{\pi}{4}} \ln(1 + \tan t) \, dt.$
利用对称性，令 $ t = \frac{\pi}{4} - u $，得
$I = \int_{0}^{\frac{\pi}{4}} \ln\left(1 + \frac{1 - \tan u}{1 + \tan u}\right) \, du = \int_{0}^{\frac{\pi}{4}} \ln\left(\frac{2}{1 + \tan u}\right) \, du.$
拆分积分得
$I = \frac{\pi}{4} \ln 2 - I,$
解得
$I = \frac{\pi}{8} \ln 2.$
答案： $\boxed{\frac{\pi}{8} \ln 2}$