#3x^2-18x+20=3x^2-18x+27-7=3(x-3)^2-7=#
#=(sqrt3(x-3))^2-7=7(((sqrt3(x-3))/sqrt7)^2-1)#
So, we have:
#I=intsqrt(7(((sqrt3(x-3))/sqrt7)^2-1))dx#
#I=sqrt7 int sqrt(((sqrt3(x-3))/sqrt7)^2-1)dx#
#t=(sqrt3(x-3))/sqrt7 => dt=(sqrt3 dx)/sqrt7 => dx=sqrt(7/3)dt#
#I=7/sqrt3 int sqrt(t^2-1)dt#
#t=secm => dt=secm tanm dm#
#t^2-1=sec^2 m-1=tan^2m#
#I=7/sqrt3 int sqrt(tan^2m) secm tanm dm#
#I=7/sqrt3 int tan^2 m secm dm = 7/sqrt3 int sin^2 m/cos^3 m dm#
#u=sinm => du=cosmdm#
#dv=sinm/cos^3m dm => v=int (dk)/k^3=-1/(2cos^2m)#
#I=7/sqrt3(-sinm/(2cos^2m) + 1/2int (cosmdm)/cos^2m)#
#I=7/sqrt3(-sinm/(2cos^2m) + 1/2int (dm)/cosm)#
#tan(m/2)=p => m=2arctanp => dm=2/(1+p^2)dp#
#p^2=(1-cosm)/(1+cosm)=2/(1+cosm)-1#
#1+cosm=2/(1+p^2) => cosm=(1-p^2)/(1+p^2)#
#I_1=int (dm)/cosm =int (2/(1+p^2))/((1-p^2)/(1+p^2))dp#
#I_1=int 2/(1-p^2)dp=int 2/((1-p)(1+p))dp#
#A/(1-p)+B/(1+p)=(A+Ap+B-Bp)/((1-p)(1+p))#
#A+B=2, A-B=0 <=> A=1, B=1#
#I_1=int (dp)/(1-p)+int (dp)/(1+p)=-ln|1-p|+ln|1+p|#
#I_1=ln|(1+p)/(1-p)|=ln|(1+tan(m/2))/(1-tan(m/2))|#
#I=7/(2sqrt3)(-sinmsec^2m+ln|(1+tan(m/2))/(1-tan(m/2))|)+C#
#m=arcsec((sqrt3(x-3))/sqrt7)#
