Question

The integral $\int \left[\left(2x-1\right)\cos \left(\sqrt{{(2x-1)}^2+5}\right)/\left(\sqrt{4x^2-4x+6}\right)\right]dx$is equal to (where $C$ is a constant of integration)

• A. $\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)\sin \left(\sqrt{{(2x+1)}^2+5}\right)+C$
• B. $\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)\sin \left(\sqrt{{(2x-1)}^2+5}\right)+C$
• C. $\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)\cos \left(\sqrt{{(2x+1)}^2+5}\right)+C$
• D. $\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)\cos \left(\sqrt{{(2x-1)}^2+5}\right)+C$

Answer 1

The correct option is B

$\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)\sin \left(\sqrt{{(2x-1)}^2+5}\right)+C$

Explanation for correct option

The given integral is $\int \left[\left(2x-1\right)\cos \left(\sqrt{{(2x-1)}^2+5}\right)/\left(\sqrt{4x^2-4x+6}\right)\right]dx$

Let ${(2x-1)}^2+5=t^2$

$\Rightarrow 2(2x-1)dx=2tdt$

$$\begin{gathered} \int \frac{\left(\cos \left(t\right)\right)}{t}\left(\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)dt \\ =\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)\sin t+C \\ =\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)\sin \sqrt{{\left(2x-1\right)}^2+5}+C \end{gathered}$$

Hence, option B is correct.