Question

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

• A. $\frac{1}{2}(x-4)\sqrt{x^2-8x+7}+9\log |x-4+\sqrt{x^2-8x+7}|+C$
• B. $\frac{1}{2}(x+4)\sqrt{x^2-8x+7}+9\log |x+4+\sqrt{x^2-8x+7}|+C$
• C. $\frac{1}{2}(x-4)\sqrt{x^2-8x+7}-3\sqrt{2}\log |x-4+\sqrt{x^2-8x+7}|+C$
• D. $\frac{1}{2}(x-4)\sqrt{x^2-8x+7}-\frac{9}{2}\log |x-4+\sqrt{x^2-8x+7}|+C$

Answer 1

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

Let $I=\int \sqrt{x^2-8x+7}dx$
$=\int \sqrt{(x^2-8x+16)-9}dx$
$=\int \sqrt{(x-4{)}^2-(3{)}^2}dx$
We know that, $\int \sqrt{x^2-a^2}dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log |x+\sqrt{x^2-a^2}|+C$
$\therefore I=\frac{(x-4)}{2}\sqrt{x^2-8x+7}-\frac{9}{2}\log |(x-4)+\sqrt{x^2-8x+7}|+C$
Hence, the correct Answer is D.