JEE Math Practice Question

The integral $$\int {{{(2x - 1)\cos \sqrt {{{(2x - 1)}^2} + 5} } \over {\sqrt {4{x^2} - 4x + 6} }}} dx$$ is equal to (where c is a constant of integration)

1. A.$${1 \over 2}\sin \sqrt {{{(2x - 1)}^2} + 5} + c$$
2. B.$${1 \over 2}\cos \sqrt {{{(2x + 1)}^2} + 5} + c$$
3. C.$${1 \over 2}\cos \sqrt {{{(2x - 1)}^2} + 5} + c$$
4. D.$${1 \over 2}\sin \sqrt {{{(2x + 1)}^2} + 5} + c$$