Question

求解不定积分 ∫dx/(x^4 (1+x^2 ) ) 我自己做出来是-1/(3*x^3)+1/x+arctanx+C

答案是-1/(3*x^3)+1/x-arctan(1/x)+C

Answer 1

[-arctan(1/x)]'=-1/(1+1/x^2)*(-1/x^2)=1/(1+x^2)
(arctanx)'=1/(1+x^2)
所以两个都对!

Answer 2

倒代换的题目x=1/t,dx=-1/t^2
∫(-1/t^2)dt/[1/t^4(1+1/t^2)]
=-∫t^4dt/(t^2+1)
=-∫(t^4-1+1)/(t^2+1)
=-∫(t^2-1)dt-∫1/(t^2+1) dt
=-t^3/3+t-arctant+C
代入t=1/x
答案是对的

Answer 3

= ∫(1/x^4 - 1/x^2 - 1/(1+x^2)) dx
= -1/(3*x^3) + 1/x + arctanx + C

But -1/(3*x^3)+1/x-arctan(1/x)+C is also correct because
( -arctan(1/x) )' = -1/(1+1/x^2) * (-1/x^2) = 1/(1+x^2) = (arctanx)'
In fact, -arctan(1/x) = arctanx - pai

Answer 4

int(1/(x^4*(1+x^2)))

ans =

(x^2 - 1/3)/x^3 - atan(1/x)