$\int \sqrt{x^2-6 x-16} \mathrm{~d} x$ equals
- A
$\left(\frac{x-3}{2}\right) \sqrt{x^2-6 x-16} +\frac{5}{2} \log \left(x-3+\sqrt{x^2-6 x-16}\right)+c$
where c is the constant of integration
- B
$$ \begin{aligned} & \left(\frac{x-3}{2}\right) \sqrt{x^2-6 x-16} -\frac{25}{2} \log \left(x-3+\sqrt{x^2-6 x-16}\right)+c \end{aligned} $$
where c is the constant of integration
- C
$\left(\frac{x-3}{2}\right) \sqrt{x^2-6 x-16}+\frac{25}{2} \log \left(x-3+\sqrt{x^2-6 x-16}\right)+c $
where c is the constant of integration
- D
$$ \begin{aligned} \left(\frac{x-3}{2}\right) \sqrt{x^2-6 x-16} & -\frac{5}{2} \log \left(x-3+\sqrt{x^2-6 x-16}\right)+c \end{aligned} $$
where $c$ is the constant of integration
✓ Correct answer: B
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