Question

求1／（X^4根号（X^2＋1））的不定积分,要详细过程.？

Answer 1

x=tant dx=sec²tdt √x²+1=sect
原式=∫{1/[(tant)^4sect]}sec²tdt
=∫(cos³t/(sint)^4)dt sint=u costdt=du
=∫[(1-u²)/u^4]du
=(-1/3)/u³+1/u+c
=(-1/3)(√(x²+1)/x)³+(√(x²+1)/x)³+c,10,