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Indefinite Integrals question

2021 · 18 Mar · Shift 1 · Q28

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JEE Main
Mathematics
Indefinite Integrals
2021 · 18 Mar · Shift 1 · Q28

Indefinite Integrals question

2021 · 18 Mar · Shift 1 · Q28

JEE MainMathematicsIndefinite IntegralsMCQ+4 / −1

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
${1 \over 2}\sin \sqrt {{{(2x + 1)}^2} + 5} + c$

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