The integral you wrote is impossible to do using "elementary functions" (at least according to the computer program Mathematica...it requires "Elliptic Functions").
I think you probably meant to do #\int(\frac{dx}{\sqrt{2x^{2}+2x+5}})#. The first steps in doing this integral is to use a bit of tricky algebra (including the technique of completing the square) to get:
#\int\frac{dx}{\sqrt{2x^{2}+2x+5}}=\frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{x^{2}+x+\frac{5}{2}}}#
#=\frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{x^{2}+x+\frac{1}{4}+\frac{9}{4}}}=\frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{(x+\frac{1}{2})^{2}+\frac{9}{4}}}#.
Now do the trigonometric substitution #x+\frac{1}{2}=\frac{3}{2}\tan(\theta)# do so that #\theta=\arctan(\frac{2}{3}x+\frac{1}{3})# and #dx=\frac{3}{2}\sec^{2}(\theta)d\theta#.
Then the integral becomes:
#\frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{(x+\frac{1}{2})^{2}+\frac{9}{4}}}=\frac{1}{\sqrt{2}}\int\frac{\frac{3}{2}\sec^{2}(\theta)d\theta}{\sqrt{\frac{9}{4}(\tan^{2}(\theta)+1)}}#, which simplifies to
#\frac{1}{\sqrt{2}}\int\frac{\sec^{2}(\theta)d\theta}{\sqrt{\sec^{2}(\theta)}}#
after using a trigonometric identity. Technically-speaking, #\sqrt{\sec^{2}(\theta)}=|sec(\theta)|#, however, it is fine to get rid of the absolute value signs for the purpose of finishing the integral (assume #\sec(\theta)\geq 0#).
Hence, the integral becomes #\frac{1}{\sqrt{2}}\int\sec(\theta)d\theta#. This last integral looks simple, but it is still tricky to do. One trick that works is to multiply it by #\frac{\sec(\theta)+\tan(\theta}}{\sec(\theta)+\tan(\theta)# and then do a substitution (you can try that if you want). Wolfram Alpha gives the answer as #\frac{1}{\sqrt{2}}(ln(cos(\theta/2)+sin(\theta/2))-\ln(\cos(\theta/2)-\sin(\theta/2)))+C#. Now just replace #\theta# by #\theta=\arctan(\frac{2}{3}x+\frac{1}{3})# and you are done:
#\int(\frac{dx}{\sqrt{2x^{2}+2x+5}})=\frac{1}{\sqrt{2}}(ln(cos(\frac{1}{2}\arctan(\frac{2}{3}x+\frac{1}{3}))+sin(\frac{1}{2}\arctan(\frac{2}{3}x+\frac{1}{3})))-\ln(\cos(\frac{1}{2}\arctan(\frac{2}{3}x+\frac{1}{3}))-\sin(\frac{1}{2}\arctan(\frac{2}{3}x+\frac{1}{3}))))+C#
It is possible to simplify this by using half angle formulas and the following equations, but it may not be worth the trouble: #\cos(\arctan(\frac{2}{3}x+\frac{1}{3}))=\frac{1}{\sqrt{1+(\frac{2}{3}x+\frac{1}{3})^{2}}}=\frac{3}{\sqrt{4x^{2}+4x+10}}#
#\sin(\arctan(\frac{2}{3}x+\frac{1}{3}))=\frac{\frac{2}{3}x+\frac{1}{3}}{\sqrt{1+(\frac{2}{3}x+\frac{1}{3})^{2}}}=\frac{2x+1}{\sqrt{4x^{2}+4x+10}}#
