Question

∫dx/x^2√(1+x^2)

Answer 1

F(x)=∫dx/[(x^2)*((1+x^2)^(1/2))]
设x=tant，则dx=sec²tdt，　
∴F(x)=∫dx/[(x^2)*((1+x^2)^(1/2))] 　
=∫sec²tdt/[tan²t*(1+tan²t)^(1/2)] 　瞎氏
=∫sec²tdt/(tan²t*sect)
=∫sectdt/tan²t
=∫(cos²t/sin²t)*(1/伏搏cost)*dt
=∫缺神祥(cost/sin²t)dt
=∫dsint/sin²t
=(-1/sint) ＋C

Answer 2

令x = tanθ，dx = sec²θdθ
∫ dx/[x²√旅握渗(1 + x²拆脊)]，1 + x² = 1 + tan²θ = sec²θ
= ∫ (sec²θ)/(tan²θsecθ) dθ
= ∫ 1/cosθ * cos²皮宏θ/sin²θ dθ
= ∫ cscθcotθ dθ
= - cscθ + C
= - √(1 + x²)/x + C