Question

不定积分求值

Answer 1

∫ dx/[x(1+x⁴)]
令u=x⁴,du=4x³ dx
原式= ∫ 1/[x*(1+u)] * du/(4x³)
= (1/4)∫ 1/[u(u+1)] du
= (1/4)∫ (u+1-u)/[u(u+1)] du
= (1/4)∫ [1/u - 1/(u+1)] du
= (1/4)(ln|u| - ln|u+1|) + C
= (1/4)ln|x^4| - (1/4)ln|x^4+1| + C
= ln|x| - (1/4)ln(x^4+1) + C
感谢 作业帮用户
这个应该是定积分，把2和1带进去减一下就行了
满意的话希望采纳哦么么哒~O(∩_∩)O

Answer 2

dx/(x(1+x^4))=xdx(x^2(1+x^4))=1/2*dx^2/(x^2(1+x^4))
=du/(2u(1+u^2))=udu/(2u^2(1+u^2))=du^2/(4u^2(1+u^2))
=dt/(4t(1+t))=(dt/t-dt/(1+t))/4
x=1 to 2==>t=u^2=x^4=1 to 16
原式=ln(t/(1+t))/4 from 1 to 16
=ln(32/17)/4