e"(1+ sin x)例10Y1 + cos xxxe*(1 + 2sin =cos22dx解?原式=2 x2cos2一t)dxtan02x22cos2x+ tan =de*}-fd(e* tan(tan122xt=+C.tan =e2
例10 d . 1 cos (1 sin ) + + x x e x x = + x x e x e x x )d 2 tan 2 2cos 1 ( 2 d ] 2 ) tan 2 [( d(tan = + x x e x x e = ) 2 d( tan x e x . 2 tan C x e x = + + x x x x e x d 2 2cos ) 2 cos 2 (1 2sin 2 解 原式=
例11 [ xarctanxn(1 + x’)dx.u[ In(1+x)d(1+x*)解：: {xln(1+x")dx=D-↓(1+x")n(1+x")-↓x*+C.2原式= Jarctanxd;(1 +x)in(1 +x)-}1uV[(1+ x)In(1+ x)-xJarctan x2(ln(1 + x2Jdx1arctan x[(1+ x)In(1+ x)-x2 -3]23xXIn(1 ++C.22
例11 arctan ln(1 )d . 2 x x + x x [(1 x )ln(1 x ) x ]arctan x 2 1 2 2 2 = + + − x x x x ]d 1 [ln(1 ) 2 1 2 2 2 + − + − u arctan [(1 )ln(1 ) 3] 2 1 2 2 2 = x + x + x − x − . 2 3 ln(1 ) 2 2 C x x x − + + + 1+ u v 解 xln(1 x )dx 2 + ln(1 )d( ) 2 1 2 2 x x = + . 2 1 (1 )ln(1 ) 2 1 2 2 2 = + x + x − x + C ] 2 1 (1 )ln(1 ) 2 1 arctan d[ 2 2 2 原式= x + x + x − x