The integral $\int {\left[ {\dfrac{{\left( {2x - 1} \right)\cos \sqrt {{{\left( {2x - 1} \right)}^2} + 5} }}{{\sqrt {4{x^2} - 4x + 6} }}} \right]} dx$ is equal to (where $c$ is constant of integration)
A. $\dfrac{1}{2}\sin \sqrt {{{\left( {2x + 1} \right)}^2} + 5} + c$
B. $\dfrac{1}{2}\sin \sqrt {{{\left( {2x - 1} \right)}^2} + 5} + c$
C. $\dfrac{1}{2}\cos \sqrt {{{\left( {2x + 1} \right)}^2} + 5} + c$
D. $\dfrac{1}{2}\cos \sqrt {{{\left( {2x - 1} \right)}^2} + 5} + c$

Hint: The expansion of the term under root in the numerator is the same as the term under root in the denominator. Also, the differentiation of the term is present in the numerator. So, make a substitution for it and use the formula of integration to find the value of the integral.

Formula Used:
$\dfrac{d}{{dx}}\left\{ {f\left( x \right) \pm g\left( x \right)} \right\} = \dfrac{d}{{dx}}\left\{ {f\left( x \right)} \right\} \pm \dfrac{d}{{dx}}\left\{ {g\left( x \right)} \right\}$, where $f\left( x \right)$ and $g\left( x \right)$ are two functions of $x$
$\dfrac{d}{{dx}}\left\{ {cf\left( x \right)} \right\} = c\dfrac{d}{{dx}}\left\{ {f\left( x \right)} \right\}$, where $c$ is a constant
$\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$, where $n$ is a real number
and the derivative of constant is zero.
$\int {\cos tdt} = \sin t + c$, where $c$ is constant of integration

Complete step by step solution:
Expand the term ${\left( {2x - 1} \right)^2} + 5$
${\left( {2x - 1} \right)^2} + 5 = 4{x^2} - 4x + 1 + 5 = 4{x^2} - 4x + 6$
Differentiate the term ${\left( {2x - 1} \right)^2} + 5$
$\dfrac{d}{{dx}}\left\{ {{{\left( {2x - 1} \right)}^2} + 5} \right\} = \dfrac{d}{{dx}}\left( {4{x^2} - 4x + 6} \right)$
 $\begin{array}{l} = 4\dfrac{d}{{dx}}\left( {{x^2}} \right) - 4\dfrac{d}{{dx}}\left( x \right) + \dfrac{d}{{dx}}\left( 6 \right)\\ = 4 \times 2x - 4 \times 1 + 0\\ = 8x - 4\\ = 4\left( {2x - 1} \right)\end{array}$
Let ${\left( {2x - 1} \right)^2} + 5 = {t^2}$
Differentiating both sides with respect to $x$, we get
$\begin{array}{l}\dfrac{d}{{dx}}\left\{ {{{\left( {2x - 1} \right)}^2} + 5} \right\} = \dfrac{d}{{dx}}\left( {{t^2}} \right)\\ \Rightarrow 4\left( {2x - 1} \right) = 2t\dfrac{{dt}}{{dx}}\\ \Rightarrow \left( {2x - 1} \right)dx = \dfrac{1}{4} \times 2tdt\\ \Rightarrow \left( {2x - 1} \right)dx = \dfrac{1}{2}tdt\end{array}$
Substituting ${\left( {2x - 1} \right)^2} + 5 = {t^2}$ and $\left( {2x - 1} \right)dx = \dfrac{1}{2}tdt$ in the given integral, we get
$\int {\left[ {\dfrac{{\left( {2x - 1} \right)\cos \sqrt {{{\left( {2x - 1} \right)}^2} + 5} }}{{\sqrt {4{x^2} - 4x + 6} }}} \right]} dx = \int {\dfrac{{\cos \sqrt {4{x^2} - 4x + 6} }}{{\sqrt {4{x^2} - 4x + 6} }}} \left( {2x - 1} \right)dx$
$\begin{array}{l} = \int {\dfrac{{\cos t}}{{{t}}} \times \dfrac{1}{2}{t}dt} \\ = \dfrac{1}{2}\int {\cos tdt} \\ = \dfrac{1}{2}\sin t + c.....(i)\end{array}$
where $c$ is constant of integration
We assumed that ${\left( {2x - 1} \right)^2} + 5 = {t^2}$
Find $t$ in terms of $x$
$\begin{array}{l}{t^2} = {\left( {2x - 1} \right)^2} + 5\\ \Rightarrow t = \sqrt {{{\left( {2x - 1} \right)}^2} + 5} \end{array}$
Substitute the expression for $t$ in $(i)$, we get
$\int {\left[ {\dfrac{{\left( {2x - 1} \right)\cos \sqrt {{{\left( {2x - 1} \right)}^2} + 5} }}{{\sqrt {4{x^2} - 4x + 6} }}} \right]} dx = \dfrac{1}{2}\sin \sqrt {{{\left( {2x - 1} \right)}^2} + 5} + c$, where $c$ is constant of integration

Option ‘B’ is correct

Note: The constant of integration must be used for indefinite integral. Differentiate the term ${\left( {2x - 1} \right)^2} + 5$ after expanding to avoid mistakes. The integration of cosine is sine but the integration of sine is (-cosine).