例5求 x 解∫ dx a+x 1+ I、a 1+ 1+ a1+ ee du==arctan+C arctan -+C. 1  2 2 + x a x 求 d  + x a x d 2 2 1    + = ax a a x d 2 1 1 1 arctan . 1 C ax a = + 例 5 解 x ax a d    + = 2 2 1 1    +   = 2 11 ax x ax a d  + = u a u d 2 1 1 1 u C a = arctan + 1