int sqrt{x^2-8 x+7} , dx = . . . . . . + C | vedclass

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Question

(D) To solve the integral $\int \sqrt{x^2-8 x+7} \, dx$,we first complete the square for the quadratic expression inside the square root:$x^2 - 8x + 7

Vedclass · Accepted Answer

$\frac{x-4}{2} \sqrt{x^2-8 x+7} - \frac{9}{2} \log \left|x-4+\sqrt{x^2-8 x+7}\right|$

Vedclass · Answer

(D) To solve the integral $\int \sqrt{x^2-8 x+7} \, dx$,we first complete the square for the quadratic expression inside the square root:$x^2 - 8x + 7 = (x^2 - 8x + 16) - 16 + 7 = (x-4)^2 - 9 = (x-4)^2 - 3^2$.Now the integral becomes $\int \sqrt{(x-4)^2 - 3^2} \, dx$.Using the standard formula $\int \sqrt{t^2 - a^2} \, dt = \frac{t}{2} \sqrt{t^2 - a^2} - \frac{a^2}{2} \log \left|t + \sqrt{t^2 - a^2}\right| + C$,where $t = x-4$ and $a = 3$:$= \frac{x-4}{2} \sqrt{(x-4)^2 - 3^2} - \frac{3^2}{2} \log \left|(x-4) + \sqrt{(x-4)^2 - 3^2}\right| + C$$= \frac{x-4}{2} \sqrt{x^2-8x+7} - \frac{9}{2} \log \left|x-4 + \sqrt{x^2-8x+7}\right| + C$.Comparing this with the given options,the correct option is $D$.

$\int \sqrt{x^2-8 x+7} \, dx = $ . . . . . . $+ C$.

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