неопр. интеграл help pls

Ok. Lets see... Oh, sorry, i dont understand this shit.

∫√(1+x²) d(1/x)=-∫√(1+x²)/x² dx
x∈(-∞,0)∪(0,+∞)
Let x=tg(v), v=arctg(x), dv=1/cos²(v)  dv and v∈(-π/2,π/2)
-∫√(1+x²)/x² dx=-∫[√(1+tg²(v))/tg²(v)] [1/cos²(v)]  dv=
=-∫√(1+tg²(v))/sin²(v) dv=
=-∫√([cos²(v)+sin²(v)]/cos²(v))/sin²(v) dv=
=-∫√(1/cos²(v))/sin²(v) dv=
=-∫1/[|cos(v)|sin²(v)] dv=
|cos(v)|=cos(v), when  v∈(-π/2,π/2)
=-∫1/[cos(v)sin²(v)] dv=
=-∫[cos²(v)+sin²(v)]/[cos(v)sin²(v)] dv=
=-∫cos²(v)/[cos(v)sin²(v)]+sin²(v)/[cos(v)sin²(v)] dv=
=-∫cos(v)/sin²(v)+1/cos(v) dv=
=-∫cos(v)/sin²(v) dv - ∫1/cos(v) dv=
=-∫1/sin²(v) d(sin(v)) -∫cos(v)/cos²(v) dv =
=1/sin(v)  - ∫1/(1-sin²(v)) d(sin(v))=
=1/sin(v)  - ∫1/(1-sin(v))(1+sin(v)) d(sin(v))=
=1/sin(v)  - ½∫1/(1-sin(v))+1/(1+sin(v)) d(sin(v))=
=1/sin(v)  + ½ln|1-sin(v)|- ½ln|1+sin(v)|+C=
=1/sin(v)  + ½ln|(1-sin(v))/(1+sin(v))|+C