⑩串紫学 Teaching Plan on Advanced Mathematics 例1求 cos 2r 解 2cos 2x=cos 2x 2dx= cos 2x(2x),'dx I cosudu =sinu+C=sin 2x+C 例2求 3+2x 解: 3+2x2J3+2t‘(3+2r) dx L=3+2x -du==lnu+c 2 ln(3+2x)+C.Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 例1 求  2cos 2xdx 解： 2cos2xdx cos2x 2dx cos2x (2x)'dx    =  =  udu u C x C u x = = + = +  = cos sin sin2 2 例2 求 dx x  3 + 2 1 x dx x (3 2 ) 3 2 1 2 1  +  + =  dx x  3 + 2 解： 1 du u u x  = + = 1 2 1 3 2 = ln u+C 2 1 ln( 3 2 ) . 2 1 = + x +C