This is hard...
#int1/sqrt(1+sqrt(1+x^2))dx#
We need to delete all this square !!
For this let's :
#u = sqrt(1+x^2)#
#x = sqrt(u^2-1)#
#du = x/sqrt(1+x^2)dx#
#dx=sqrt(1+x^2)/xdu#
Integral become :
#intsqrt(1+x^2)/(x*sqrt(1+sqrt(1+x^2)))du#
#intu/(sqrt(u^2-1)*sqrt(u+1))du#
Multiply numerator and denominator by#sqrt(u-1)# and don't forget that #(u+1)(u-1) = u^2-1# so we have :
#(usqrt(u-1))/(u^2-1)#
Let's #w = sqrt(u-1)#
#dw = 1/(2sqrt(u-1))du#
#du = 2sqrt(u-1)dw#
#2int(usqrt(u-1))/(u^2-1)sqrt(u-1)dw#
#w^2+1=u#
#(w^2+1)^2=u^2#
#2int((w^2+1)w^2)/((w^2+1)^2-1)dw#
#2int(w^4+w^2)/(w^4+2w^2)dw#
#2int(w^2(w^2+1))/(w^2(w^2+2))dw#
#2int(w^2+1)/(w^2+2)dw#
#2int(w^2+2-1)/(w^2+2)dw#
#2int1dw-2int1/(w^2+2)dw#
#2int1dw-int1/(1/2w^2+1)dw#
Let's #t =1/sqrt(2)w#
#t^2=1/2w^2#
#dt = 1/sqrt(2)dw#
#[2w]- sqrt(2)int1/(t^2+1)dt#
#[2w]-[sqrt(2)arctan(t)]+C#
Substitute back...
#[2w]-[sqrt(2)arctan(1/sqrt(2)w)]+C#
#[2sqrt(u-1)]-[sqrt(2)arctan(1/sqrt(2)sqrt(u-1))]+C#
#[2sqrt(sqrt(x^2+1)-1)]-[sqrt(2)arctan(1/sqrt(2)sqrt(sqrt(x^2+1)-1))]+C#
