Something Mathematical to do.

I used this site:

https://goo.gl/Qem530

and got this answer:

I haven't tried to make sense of it yet though.

Something Mathematical to do.

\(\begin{array}{llcl} \int \limits_{x=0}^{\infty} { \frac{1}{ (x+\sqrt{1+x^2})^2 } \ dx}\\\\ & \text{substitute:}\\ & \boxed{~ x=\tan(z) \\ dx = ( 1+\tan^2(z))\ dz \\ ~}\\ & \text{new limits:}\\ & \boxed{~ z=\arctan(0) \Rightarrow z = 0 \\ z=\arctan(\infty) \Rightarrow z = \frac{\pi}{2} \\\\ ~}\\ =\int \limits_{z=0}^{\frac{\pi}{2}} { \frac{1+\tan^2(z)}{ \Big(\tan(z)+\sqrt{1+\tan^2(z)}\Big)^2 } \ dz} \\ & \boxed{~ 1+\tan^2(z) = \frac{1}{\cos^2(z)} \\ ~}\\ =\int \limits_{z=0}^{\frac{\pi}{2}} { \frac{1}{\cos^2(z)}\cdot \left( \frac{1}{\frac{1}{\cos^2(z)}+ \tan^2(z)+2\cdot \frac{\sin(z)}{\cos^2(z)} } \right) \ dz} \\ = \int \limits_{z=0}^{\frac{\pi}{2}} { \frac{1}{1+2\cdot \sin(z) + \sin^2(z) } \ dz} \\ = \int \limits_{z=0}^{\frac{\pi}{2}} { \frac{1}{ \Big(1+\sin(z)\Big)^2 } \ dz} \\ & \text{substitute:}\\ & \boxed{~ t=\tan(\frac{z}{2}) \\ dt = \frac12 \cdot \left( 1+\tan^2(\frac{z}{2}) \right)\ dz\\ =\frac12 \cdot (1+t^2)\ dz \\ ~}\\ & \text{new limits:}\\ & \boxed{~ t=\tan(\frac{0}{2}) \Rightarrow t = 0 \\ t=\tan(\frac{ \frac{\pi}{2}}{2}) \Rightarrow t = 1 \\\\ ~}\\ = \int \limits_{t=0}^{1} \frac{2\ dt}{1+t^2} \cdot \left( \frac{1}{ \left(1+ \sin(z)\right)^2 } \right) \\ & \boxed{~ \sin(z) = \frac{2t}{1+t^2} \\ ~}\\ = \int \limits_{t=0}^{1} \frac{2\ dt}{1+t^2} \cdot \left( \frac{1}{ \left(1+ \frac{2t}{1+t^2}\right)^2 } \right) \\ = 2\cdot \int \limits_{t=0}^{1} \frac{1+t^2}{1+4t+6t^2+4t^3+t^4} \ dt\\ = 2\cdot \int \limits_{t=0}^{1} \frac{1+t^2}{(1+t)^4} \ dt\\ \end{array}\)

\(\begin{array}{llcl} & \text{Partial fraction decomposition:}\\ & \boxed{~ \begin{array}{lcll} & \frac{1+t^2}{(1+t)^4} &=& \frac{A}{1+t} + \frac{B}{(1+t)^2} + \frac{C}{(1+t)^3} + \frac{D}{(1+t)^4} \\\\ & 1+t^2 &=& A\cdot (1+t)^3 +B\cdot (1+t)^2 +C\cdot (1+t) +D \\ (1)\quad t=-1 : & \Rightarrow 2 &=& D \\ (2)\quad t= 0 : & \Rightarrow -1 &=& A+B+C \\ (3)\quad t= 1 : & \Rightarrow 0 &=& 8A+4B+2C \quad | \quad : 2 \\ & 0 &=& 4A+2B+C \\ (4)\quad t= -2 : & \Rightarrow 3 &=& -A+B-C \\\\ (2)+(4) : & -1+3 &=& 2 B \qquad \Rightarrow B =1 \\ (3)+(4) : & 0+3 &=& 3A + 3B \quad | \quad : 3 \\ & 1 &=& A + B \quad | \quad B=1 \\ & 1 &=& A + 1 \qquad \Rightarrow A =0 \\ (2): & C &=& -1-A-B \\ & &=& -1-0-1 \qquad \Rightarrow C =-2 \\ \end{array} ~}\\ \end{array}\)

\(\begin{array}{llcl} = 2 \int \limits_{t=0}^{1} \left( \frac{1}{(1+t)^2} -2\cdot \frac{1}{(1+t)^3} +2\cdot \frac{1}{(1+t)^4 } \right) \ dt\\ = 2 \int \limits_{t=0}^{1} \frac{1}{(1+t)^2}\ dt -4 \int \limits_{t=0}^{1} \frac{1}{(1+t)^3}\ dt +4 \int \limits_{t=0}^{1} \frac{1}{(1+t)^4}\ dt \\ = 2 \left[ -\frac{1}{1+t} \right]_{t=0}^{1} -\frac{4}{2} \left[ -\frac{1}{(1+t)^2} \right]_{t=0}^{1} +\frac{4}{3} \left[ -\frac{1}{(1+t)^3} \right]_{t=0}^{1} \\ = 2 \left[ -\frac{1}{2} -(-1) \right] -\frac{4}{2} \left[ -\frac{1}{4}-(-1) \right] +\frac{4}{3} \left[-\frac{1}{8}-(-1) \right] \\ = 2 \left[ -\frac{1}{2} +1 \right] -2 \left[ -\frac{1}{4}+1 \right] +\frac{4}{3} \left[-\frac{1}{8}+1 \right] \\ = 2 \cdot \frac12 -2\cdot \frac34 +\frac43 \cdot \frac78 \\ = 1 - \frac32+ \frac76 \\ = \frac66 - \frac96 +\frac76 \\ = \frac{13}{6} - \frac96 \\ = \frac46 \\ = \frac23 \\ \end{array}\)

\(\begin{array}{llcl} \int \limits_{x=0}^{\infty} { \frac{1}{ (x+\sqrt{1+x^2})^2 } \ dx} = \frac23 \end{array}\)

Thanks Heureka,

I was wondering how this would be done.  

I did think of substituting x=tan(z) but I didn't get very far with it.   

Now I shall take a good look at your answer :))