例6求积分∫ dx x 1+e2+e3+ 解令t=e→x=6nt,d 6 dt d dt 1+t3+t2+t 1+e2+e3+e 6 dt 633t+3 dt r(1+t)(1+ t1+t1+t2
例6 求积分 解 . 1 1 2 3 6 dx e e e x x x + + + 令 6 x t = e x = 6lnt, , 6 dt t dx = dx e e e x x x + + + 2 3 6 1 1 dt t t t t 6 1 1 3 2 + + + = dt t t t + + = (1 )(1 ) 1 6 2 dt t t t t + + − + = − 2 1 3 3 1 6 3
633t+3 dt t1+t1+t2 3a(1+t2) 6Int-3In(1+t) 3 dt 2J1+ 1+t2 3 6Int-3In(1+t)-In(+t)-3arctant+C 2 3 x-3In(1+e6)-In1 +e3)-3arctan(e6)+C 2
= t − + t − ln(1+ t ) − 3arctant + C 2 3 6ln 3ln(1 ) 2 dt t t t t + + − + = − 2 1 3 3 1 6 3 ln(1 ) 3arctan( ) . 2 3 3ln(1 ) x e 6 e 3 e 6 C x x x = − + − + − + 2 3 = 6lnt − 3ln(1+ t) − dt t t d t + − + + 2 2 2 1 1 3 1 (1 )