Calculus: How do I integrate square root (x^2 +2x) ?

sqrt(x^2+2x)=sqrt((x+1)^2-1)

y=(x+1)^2
dy = 2(x+1) dx => dy/2*sqrt(y) = dx

2* Int. sqrt(y-1)/sqrt(y) dy
=2*Int. sqrt(1-1/y) dy

1-1/y = z^2 => 1/y^2 dy = 2zdz

dy = 2zdz/(1-z^2)^2

=4*Int. z^2/(1-z^2)^2 dz

=4*Int. ((z^2-1)+1)/(1-z^2)^2 dz

=4*Int 1/(z^2-1) dz + 4* Int. 1/(z^2-1)^2 dz

=2*Int dz/(z-1) - 2*Int. dz/(z+1) + 4*I ------- (1)

I = Int. 1/(z-1)^2*(z+1)^2  dz

Repeatedly substitute 1 = ((z+1)-(z-1))/2

you will get

I = (1/4)* Int. dz/(z-1) - (1/4)* Int. dz/(z+1) - (1/4)*Int. dz/(z+1)^2
-(1/4)*Int. dz/(z-1)^2

The final total Integral  from (1) evaluates to

=3* Int. dz/(z-1) - 3*Int. dz/(z+1) - Int. dz/(z+1)^2 - Int. dz/(z-1)^2
=3 ln(z-1) - 3 ln(z+1)+1/(z+1)+1/(z-1)
=3 ln((z-1)/(z+1))+2z/(z^2-1)
Substitute z= sqrt(1-1/y)
=3ln((sqrt(1-1/y)-1)/(sqrt(1-1/y)+1)) -2ysqrt(1-1/y)
substitute y=(x+1)^2
=3ln((sqrt(x^2+2x-1)-sqrt(x^2+2x))/(sqrt(x^2+2x-1)+sqrt(x^2+2x)))
 -2(x+1)^2sqrt((x^2+2x-1)/(x^2+2x)) + C