__dx_ ∫ （√a+b/x ）怎么求积分?∫dx/[（a+b/x）^05] 这个。

∫1/√(a+b/x) dx
令u=1/x,du=(-1/x²)dx,x=1/u
=∫-1/[u²√(a+bu)] du
=-∫1/[u²√(a+bu) du
令t=a+bu,dt=bdu,u=(t-a)/b
=-b²∫1/[(t-a)²√t]*1/b dt
=-b∫1/[(t-a)²√t] dt
令w=√t,t=w²,dt=2wdw
=-2b∫1/[(w²-a)²w]*w dw
=-2b∫1/(w²-a)² dw
令w=√a*secθ,dw=√a*secθtanθdθ
=-2b√a∫secθtanθ/(asec²θ-a)² dθ
=-2b√a∫secθtanθ/(a²tan^4θ) dθ
=-2b/a^(3/2)*∫(1/cosθ*cos³θ/sin³θ) dθ
=-2b/a^(3/2)*∫cot²θcscθdθ
=-2b/a^(3/2)*∫(csc²θ-1)cscθdθ
=-2b/a^(3/2)*∫(csc³θ-cscθ)dθ
=-2b/a^(3/2)*[∫csc³θdθ-∫cscθdθ]
=-2b/a^(3/2)*[-(1/2)cosθcsc²θ+(1/2)∫cscθdθ-∫cscθdθ]
=-2b/a^(3/2)*[-(1/2)cosθcsc²θ-(1/2)∫cscθdθ]
=-2b/a^(3/2)*[-(1/2)cosθcsc²θ+(1/2)ln|cotθ+cscθ|+C
之后再代回所有数值即可.
化简为
(x/a)√(a+b/x)-{1/[2a^(3/2)]*[bln|2x√a*√(a+b/x)+2ax+b|}+C