例5求「 rarctan xIn(1+x2)dh 解 ∵xln(1+x (1+x2)d(1+x2 (1+x2)ln(1+x2)-x2+C. 2 原式=∫ arctan xd(1+x)In(1+re)1 2 C1+xIn(1+x-xarctan x ln(1+x2) Jdx 1+x例 5 解 arctan ln(1 ) . 2  求 x x + x dx xln(1 x )dx 2   + ln(1 ) (1 ) 21 2 2 = + x d + x  . 21 (1 )ln(1 ) 21 2 2 2 = + x + x − x + C ] 21 (`1 )ln(1 ) 21 arctan [ 2 2 2 = xd + x + x − x 原 式  [(`1 x )ln(1 x ) x ]arctan x 21 2 2 2 = + + − dx x x x ] 1 [ln(1 ) 21 2 2 2 + − + − 