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$\int \sqrt{x^2-8 x+7} d x$ is equal to
(a) $\frac{1}{2}(x-4) \sqrt{x^2-8 x+7}+\log \left|x-4+\sqrt{x^2-8 x+7}\right|+C$
(b) $\frac{1}{2}(x+4) \sqrt{x^2-8 x+7}+9 \log \left|x+4+\sqrt{x^2-8 x+7}\right|+C$
(c) $\frac{1}{2}(x-4) \sqrt{x^2-8 x+7}-3 \sqrt{2} \log \left|x-4+\sqrt{x^2-8 x+7}\right|+C$
(d) $\frac{1}{2}(x-4) \sqrt{x^2-8 x+7}-\frac{9}{2} \log \left|x-4+\sqrt{x^2-8 x+7}\right|+C$
$\int \sqrt{x^2-8 x+7} d x$ is equal to
(a) $\frac{1}{2}(x-4) \sqrt{x^2-8 x+7}+\log \left|x-4+\sqrt{x^2-8 x+7}\right|+C$
(b) $\frac{1}{2}(x+4) \sqrt{x^2-8 x+7}+9 \log \left|x+4+\sqrt{x^2-8 x+7}\right|+C$
(c) $\frac{1}{2}(x-4) \sqrt{x^2-8 x+7}-3 \sqrt{2} \log \left|x-4+\sqrt{x^2-8 x+7}\right|+C$
(d) $\frac{1}{2}(x-4) \sqrt{x^2-8 x+7}-\frac{9}{2} \log \left|x-4+\sqrt{x^2-8 x+7}\right|+C$
Solution:
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Verified Answer
(d) $\int \sqrt{x^2-8 x+7} d x=\int \sqrt{(x-4)^2-3^2} d x$
$=\frac{x-4}{2} \sqrt{x^2-8 x+7}-\frac{9}{2} \log \left|(x-4)+\sqrt{x^2+8 x+7}\right|+C$
$=\frac{x-4}{2} \sqrt{x^2-8 x+7}-\frac{9}{2} \log \left|(x-4)+\sqrt{x^2+8 x+7}\right|+C$
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