⇒∫1−x2+6x−7dx=∫1−x2−6x+7dx⇒∫1−(x−3)2+7−9dx =∫1−(x−3)2−(2)2dx⇒∫1(2)2−(x−3)2dx=122log⁡2+x−32−x+3+c

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$\int \frac{1}{6 x - 7 - x^{2}} dx =$

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$\frac{1}{2 \sqrt{2}} \log |\frac{\sqrt{2} + x - 3}{\sqrt{2} - x + 3}| + c$

$\frac{1}{2 \sqrt{2}} \log |\frac{\sqrt{2} + x + 3}{\sqrt{2} + x + 3}| + c$

$\frac{1}{\sqrt{2}} \log |\frac{\sqrt{2} + x - 3}{\sqrt{2} - x + 3}| + c$

$\frac{1}{\sqrt{2}} \log |\frac{\sqrt{2} - x - 3}{\sqrt{2} - x + 3}| + c$

answer is C.

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Detailed Solution

$\begin{matrix} \Rightarrow \int \frac{1}{- x^{2} + 6 x - 7} d x \\ = \int \frac{1}{- [x^{2} - 6 x + 7]} d x \Rightarrow \int \frac{1}{- [(x - 3)^{2} + 7 - 9]} d x \\ = \int \frac{1}{- [(x - 3)^{2} - (\sqrt{2})^{2}]} d x \Rightarrow \int \frac{1}{(\sqrt{2})^{2} - (x - 3)^{2}} d x \\ = \frac{1}{2 \sqrt{2}} \log |\frac{\sqrt{2} + x - 3}{\sqrt{2} - x + 3}| + c \end{matrix}$