Since it is easier to deal with only one #x# under a square root, we complete the square:
#x^2+4x=(x+2)^2+k#
#x^2+4x=x^2+4x+4+k#
#k=-4#
#x^2+4x=(x+2)^2-4#
#int\ sqrt(x^2+4x)\ dx=int\ sqrt((x+2)^2-4)\ dx#
Now we need to do a trigonometric substitution. I'm going to use hyperbolic trig functions (because secant integral usually aren't very nice). We want to use the following identity:
#cosh^2(theta)-1=sinh^2(theta)#
To do this, we want #(x+2)^2=4cosh^2(theta)#. We can solve for #x# to get what substitution we need:
#x+2=2cosh(theta)#
#x=2cosh(theta)-2#
To integrate with respect to #theta#, we have to multiply by the derivative of #x# with respect to #theta#:
#dx/(d theta)=2sinh(theta)#
#int\ sqrt((x+2)^2-4)\ dx=int\ sqrt((2cosh(theta))^2-4)*2sinh(theta)\ d theta=#
#=2int\ sqrt(4cosh^2(theta)-4)*sinh(theta)\ d theta=2int\ sqrt(4(cosh^2(theta)-1))*sinh(theta)\ d theta=#
#=2*sqrt(4)int\ sqrt(cosh^2(theta)-1)*sinh(theta)\ d theta=#
Now we can use the identity #cosh^2(theta)-1=sinh^2(theta)#:
#=4int\ sqrt(sinh^2(theta))*sinh(theta)\ d theta=4int\ sinh^2(theta)\ d theta#
Now we use the identity:
#sinh^2(theta)=1/2(cosh(2theta)-1)#
#4/2int\ cosh(2theta)-1\ d theta=int\ 2cosh(2theta)\ d theta-2theta=#
We could do an explicit u-substitution for #2cosh(2theta)#, but it is pretty obvious that the answer is #sinh(2theta)#:
#=sinh(2theta)-2theta+C#
Now we need to undo the substitution. We can solve for #theta# to get:
#theta=cosh^-1((x+2)/2)#
This gives:
#sinh(2cosh^-1((x+2)/2))-2cosh^-1((x+2)/2)+C#
