解 ∫x
(sinax eb1x1)dx=∫sinaxd(ax) b∫ebd()= cosax beb+C. abaxx
(6)∫sindt;
解 ∫sint
tdt=2∫sind= 2cos+C.
(7)∫tan10x sec2xdx;
解 ∫tan10x sec2xdx=∫tan10xdtanx=
(8)∫
解 dx; xlnxlnlnx1tan11x+C. 11dx11=dlnx=∫xlnxlnlnx∫lnxlnlnx∫lnlnxdlnlnx=ln|lnlnx|+C.
x
+x2 (9)∫tan+x2 dx;
解 ∫tan+x 2x
+x2dx=∫tan+xd+x=∫22sin+x2cos+x2+x2
= ∫
(10)∫1cos+x2cos+x2= ln|cos+x2|+C. dx; sinxcosx
dxsec2x1 解 ∫=∫dx=∫dtanx=ln|tanx|+C. sinxcosxtanxtanx
1 (11)∫x xdx; e+e
解
(12)∫xe xdx;
解 ∫xe xdx= 2∫ex+e x21ex1dx=∫2xdx=∫x=arctanex+C. 2xe+11+e1 x21 x22ed( x)= e+C. 2∫2