Question

求不定积分∫√(x^2-a^2)/x^4 dx（a>0)

Answer 1

设x=asect,dx=asect*tantdt,
cost=a/x,sint=√（x^2-a^2)/x,
原式=∫a√[sect)^2-1]*asect*tantdt/[a^4*(sect)^4]
=(1/a^2)∫(tant)^2dt/(sect)^3
=(1/a^2)∫[(sint)^2/(cost)^2]*(cost)^3dt
=(1/a^2)∫（sint)^2*costdt
=(1/a^2)∫（sint)^2d(sint)
=(1/a^2)(sint)^3/3+C
=(1/a^2)*[√（x^2-a^2)/x]^3/3+C
=1/(3a^2x^3)(x^2-a^2)^(3/2)+C.

Answer 2

令 x=asect
√(x^2-a^2)=atant
dx=asecttantdt
∫√(x^2-a^2)/x^4 dx=∫{[a^2sect(tant)^2]/a^4(secx)^4}dt
=(1/a^2)∫(tant)^2(cost)^3dt=(1/a^2)∫[(sect)^2-1](cost)^3dt
=(1/a^2)[∫cost)dt-∫(cost)^3dt]=(1/a^2)[∫cost)dt-∫(1-(sint)^2)dsint]
=(1/3a^2)(sint)^3+c=(1/3a^2)(√(x^2-a^2)/x)^3+c

Answer 3

(-a^2+x^2 )^(3⁄2)/(3*a^2 *x^3 )+C