=x2sinx+2xcosx 2∫cosxdx=x2sinx+2xcosx 2sinx+C.
12. ∫te 2tdt;
解 ∫te 2tdt= 11 2t1 2t 2ttde= te+∫edt 2∫22
1111 = te 2t e 2t+C= e 2t(t++C. 2422
13. ∫ln2xdx;
1 解 ∫ln2xdx=xln2x ∫x 2lnx dx=xln2x 2∫lnxdx x
1 =xln2x 2xlnx+2∫x dx=xln2x 2xlnx+2x+C. x
14. ∫xsinxcosxdx;
解 ∫xsinxcosxdx=1111xsin2xdx= xdcos2x= xcos2x+cos2xdx 2∫4∫44∫
11 = xcos2x+sin2x+C. 48
15. ∫x2cos2
解 ∫x2cos2xdx; 2x11111dx=∫x2(1+cosx)dx=x3+∫x2dsinx=x3+x2sinx ∫xsinxdx 226262
1111 =x3+x2sinx+∫xdcosx=x3+x2sinx+xcosx ∫cosxdx 6262
11 =x3+x2sinx+xcosx sinx+C. 62
16. ∫xln(x 1)dx;
解 ∫xln(x 1)dx=1121212ln(x 1)dx=xln(x 1) x dx 2∫22∫x 1
111 =x2ln(x 1) ∫(x+1+ dx 22x 1
1111 =x2ln(x 1) x2 x ln(x 1)+C. 2422
17. ∫(x2 1)sin2xdx;