(37)∫x2 9dx; x
x2 9令x=3sect9sec2t 9(3sect)=3∫tan2tdt dx∫3sectx 解 ∫
=3∫(321)3tan393arccos dt=t t+C=x +C. 2xcost1
(38
)∫
解 dx1+2xdx
2x; 令2x=t11= tdt(1∫1+t∫1+tdt=t ln(1+t)+C=2x ln(1+2x)+C. ∫1+
(39)∫dx1+ x
dx2; 解 ∫令x=sint
1+ x21112tcostdt=(1 )dt=(1 secdt ∫1+cost∫1+cost∫22
txsint+C=arcsinx +C. =t tan+C=t 221+cost1+ x
(40)∫dx
x+ x
dx
x+ x22. 解 ∫令x=sint∫sint+cost costdt=2∫11cost+sint+cost sintdt sint+cost