Question

不定积分计算题，求个解答过程，谢谢

Answer 1

解：设F(x)=cosx/x是f(x)的一个原函数,则f(x)=(-x*sinx-cosx)/x^2,
f(x/2)=[(-x/2)*sin(x/2)-cos(x/2)]/(x^2/4)
∫xf’(x/2)dx=2∫xdf(x/2)
=2xf(x/2)-2∫f(x/2)dx
=2xf(x/2)-4∫f(x/2)d(x/2)
=2xf(x/2)-4F(x/2)+C
=-8[(x/2)*sin(x/2)+cos(x/2)]/x-4cos(x/2)/(x/2)+C(C为常数)
=-8［(x/2)*sin(x/2)+cos(x/2)/x］/x-8cos(x/2)/x+C
我观察(随意看一下)各者答案似乎不同,我的答案还能化简一下,请自己继续化简一下,不求采纳，只愿你能明白！

Answer 2

Answer 3

追答

如有帮助请采纳，谢谢

Answer 4