∫(X^4/X^2 -1)dx

∫(X^4/X^2 -1)dx=∫[(X^4+1-1)/X^2 -1]dx=∫(X^2+1+1/(X^2 -1))dx
=1/3X^3+x+1/2∫(1/X -1 -1/X+1)dx=1/3X^3+x+1/2ln(x-1)-1/2ln(x+1)+C

X^4/（X^2 -1）=（X^4-1+1）/（X^2 -1）=X^2+1+1/(X^2-1)
=X^2+1+1/[(X-1)(X+1)]=X^2+1+[1/(X-1)-1/(X+1)]/2
记住最后加个积分常数

积分:x^4/(x^2-1)dx
=积分:(x^4-1+1)dx/(x^2-1)(分解,跟刚才的那道题类似)
=积分:[(x^2+1)(x^2-1)+1]dx/(x^2-1)
=积分:(x^2+1)dx+积分:dx/(x^2-1)
=1/3*x^3+x+1/2*ln|(x-1)/(x+1)|+C
(C 是积分常数)

这里用到公式:
积分:dx/(x^2-a^2)
=1/2a*ln|(x-a)/(x+a)|+C

X^4/X^2 -1
=(X^4-1)/(X^2 -1)+1/(X^2 -1)
=X^2+1 + 1/[(X+1)(X -1)]
=X^2+1 + (1/2)*[1/(X-1) - 1/(X+1)]

所以
∫(X^4/X^2 -1)dx
=1/3 x^3 + x + 1/2 ln((x-1)/(x+1)) +C

分目变成（x^2-1）^2+2*（x^2-1）+1，所以，原式变成x^2+1+1/(x^2-1)的积分是x^3/3+x+sinx