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)
- A$${1 \over 2}\sin \sqrt {{{(2x - 1)}^2} + 5} + c$$
- B$${1 \over 2}\cos \sqrt {{{(2x + 1)}^2} + 5} + c$$
- C$${1 \over 2}\cos \sqrt {{{(2x - 1)}^2} + 5} + c$$
- D$${1 \over 2}\sin \sqrt {{{(2x + 1)}^2} + 5} + c$$
✓ Correct answer: A
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