x^ 2 -8 x+7 d x is equal to - Vidyadip

Question

$\int \sqrt{x^{2}-8 x+7} d x$ is equal to

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Answer

$I=\int \sqrt{x^{2}-8 x+7} d x$
$=\int \sqrt{\left(x^{2}-8 x+16\right)-9} d x$
$=\int \sqrt{(x-4)^{2}-(3)^{2}} d x$
We know that
$\Rightarrow \int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log |x+\sqrt{x^{2}-a^{2}}|+C$
Therefore,
$\Rightarrow I=\frac{(x-4)}{2} \sqrt{x^{2}-8 x+7}-\frac{9}{2} \log |(x-4)+\sqrt{x^{2}-8 x+7}|+C$

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