刚s.求1n=「+a dx 分部积分法对应题型 -2nx 解令w+am-1则 (x2+a2)+,v=x +a )dr X +2nln-2ndl 得递推公式In+1= 1 2na2 (x2+a2ym 2n-In 2na 例5. 求 解: 令 , ( ) 1 2 2 n x a u + = v  =1, 则 , ( ) 2 2 2 +1 + −  = n x a nx u v = x n  I x x a x n n d ( ) 2 2 2 1 2  + + + n x a x ( ) 2 2 + = x x a n n d ( ) 2  2 2 +1 + + n x a x ( ) 2 2 + = n + 2n I 1 2 − 2 n+ na I 得递推公式 n n n I na n x a x na I 1 2 2 2 2 2 2 1 2 ( ) 1 − + + + = 2 2 2 (x + a ) − a n x a x ( ) 2 2 + = 分部积分法对应题型