√(1+x)=t
dt/dx=(1/2)/√(1+x)=t/2
y=2t+log|(t-1)/(t+1)|
dy/dx=2*dt/dx+|(t+1)/(t-1)|*(d/dx)((t-1)/(t+1))
=t+|(t+1)/(t-1)|*(d/dt)((t-1)/(t+1))*dt/dx
=t+|(t+1)/(t-1)|*((t+1)-(t-1))/(t+1)^2 *(t/2)
=t+|(t+1)/(t-1)|*t/(t+1)^2
=t+t/((t-1)(t+1))
=t+t/(t^2-1)
=t(1+1/(t^2-1))
=t(ｔ^2/(t^2-1))
=ｔ^3/(t^2-1)
=(1+x)√(1+x)/x

1/(√(1+x))+1/(x√(1+x))
=(1+x)/(x√(1+x))
=(1+x)√(1+x)/(x(1+x))
=√(1+x)/x

>>652
1/x=nπ+ｔ