int sqrt{x^{2}-8 x+7} , dx is equal to | vedclass

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Question

(A) Let $I = \int \sqrt{x^{2}-8 x+7} \, dx$. Completing the square,we have $x^{2}-8x+7 = (x-4)^{2}-16+7 = (x-4)^{2}-3^{2}$. Thus,$I = \int \sqrt{(x-4)

Vedclass · Accepted Answer

$\frac{1}{2}(x-4) \sqrt{x^{2}-8 x+7}-\frac{9}{2} \log |x-4+\sqrt{x^{2}-8 x+7}|+C$

Vedclass · Answer

(A) Let $I = \int \sqrt{x^{2}-8 x+7} \, dx$. Completing the square,we have $x^{2}-8x+7 = (x-4)^{2}-16+7 = (x-4)^{2}-3^{2}$. Thus,$I = \int \sqrt{(x-4)^{2}-3^{2}} \, dx$. Using the standard formula $\int \sqrt{x^{2}-a^{2}} \, dx = \frac{x}{2} \sqrt{x^{2}-a^{2}} - \frac{a^{2}}{2} \log |x + \sqrt{x^{2}-a^{2}}| + C$,where $x$ is replaced by $(x-4)$ and $a=3$: $I = \frac{(x-4)}{2} \sqrt{(x-4)^{2}-3^{2}} - \frac{3^{2}}{2} \log |(x-4) + \sqrt{(x-4)^{2}-3^{2}}| + C$. Simplifying the expression,we get $I = \frac{1}{2}(x-4) \sqrt{x^{2}-8x+7} - \frac{9}{2} \log |x-4+\sqrt{x^{2}-8x+7}| + C$. Therefore,the correct option is $A$.

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

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