Question

∫dx/[x^4√(1+x^2)]求不定积分

Answer 1

令x= tant,dx=secx^2dt
原式=∫sect^2/（tant^4+√tant^2 +1) dt
=∫(sect/ tant^4) dt
=∫csct*cott dt
=∫csct*(csct^2-1)* cot dt
=∫csct^2-1 dcsct
= csc-（csc^3/3)+C
其中t= arctanx,所以csct=（√1+ x^2)/ x
结果为(√1+ x^2/ x)-[(√1+x^2)^3)/3]+ C

Answer 2

引用huamin8000的回答：
x=tant
∫dx/[x⁴√(1+x²)]=∫dtant/[tan⁴t√(1+tan²t)]
= ∫sect/tan⁴tdsint=∫cos³t/sin⁴tdt
=∫cos²t/sin⁴tdsint=∫1 /sin⁴ t-1/sin⁴tdsint
=-1/sint+1/(3sin³t)+C
=-sect/tant+sec³t/(3tan³t)+C
=-√(1+x²)/x+√(1+x²)³/(3x³)+C

x=tant dx=sec²tdt
∫dx/[x⁴√(1+x²)]=∫sec²t/[tan⁴t√(1+tan²t)]dt
= ∫sect/tan⁴tdt=∫cos³t/sin⁴tdt
=∫cos²t/sin⁴tdsint=∫1 /sin⁴ t-1/sin²tdsint
=1/sint-1/(3sin³t)+C
=sect/tant-sec³t/(3tan³t)+C
=√(1+x²)/x-√(1+x²)³/(3x³)+C