Question

高数题，求解

Answer 1

这道题目首先分子分母同乘以x的四次方，然后换元法求解，最后不要忘了反代回去，希望对你有帮助

Answer 2

let
x^5 = tanu
5x^4. dx = (secu)^2 du
∫ dx/[ x(1+x^10)^2]
=(1/5)∫ 5x^4dx/[ x^5.(1+x^10)^2]
=(1/5)∫ (secu)^2 du/[ tanu. (secu)^2 ]
=(1/5)∫ (cosu/sinu) du
=(1/5) ln|sinu| + C
=(1/5) ln|x^5/√(1+x^10)| + C
=ln|x| -(2/5)ln|1+x^10| + C

Answer 3

原式=∫x^4/[x^5(1+x^10) dx = 1/5 ∫1/[x^5(1+x^10)] dx^5
=1/5 ∫dt/t(1+t^2)
=1/5 dt [a/t +(b+ct)/(1+t^2)
所以a+at^2+bt +ct^2 = 1, a =1, b=0,c=-1
原来积分=1/5∫1/t dt -1/5 ∫t/(1+t^2) dt
=1/5 lnt -1/10 ln(1+t^2)+C
=1/5 lnx^5 -1/10 ln(1+x^10)+C
=1/10 ln[x^10/(1+x^10)] +C