Writing this "pretty", you want to integrate . (I presume you "completed the square" to get that denominator.)
Yes, you can make the substitution u= x- 2.5 so that du= dx and x= u+ 2.5. That makes the integral
For the second of those integrals you are asking "Can i return the substitution of t back as 2.5 * Integral of dx/[(x-2.5)^2-6.25]?"
You could but there is no point! It is simpler to do as is:
Use "partial fractions".