A cura di: Gianluca

$int1/((sqrt(x))(sqrt(1-x)))dx$

Pongo $sqrt(1-x)=t$ $rarr$$1-x=t^2$ $rarr$ $x=1-t^2$  dx=2tdt

$int1/(sqrt(1-t^2))[1/t(2t)dt]$

=$2int1/(sqrt(1-t^2))dt$

=$2arcsensqrt(x)+C$

 

di Anoè Gianluca – www.chenesoio.com