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Question

Mathematics Question on integral

∫x2−8x+7dx\int \sqrt{x^2-8x+7}dx is equal to

A

12(x−4)x2−8x+7+9log⁡∣x−4+x2−8x+7∣+C\frac{1}{2}(x-4)\sqrt{x^2-8x+7}+9\log\mid x-4+\sqrt{x^2-8x+7}\mid+C

B

12(x+4)x2−8x+7+9log⁡∣x+4+x2−8x+7∣+C\frac{1}{2}(x+4)\sqrt{x^2-8x+7}+9\log\mid x+4+\sqrt{x^2-8x+7}\mid+C

C

12(x−4)x2−8x+7−32log⁡∣x−4+x2−8x+7∣+C\frac{1}{2}(x-4)\sqrt{x^2-8x+7}-3\sqrt 2\log\mid x-4+\sqrt{x^2-8x+7}\mid+C

D

12(x−4)x2−8x+7−92log∣x−4+x2−8x+7∣+C\frac{1}{2}(x-4)\sqrt{x^2-8x+7}-\frac{9}{2}log\mid x-4+\sqrt{x^2-8x+7}\mid+C

Answer

12(x−4)x2−8x+7−92log∣x−4+x2−8x+7∣+C\frac{1}{2}(x-4)\sqrt{x^2-8x+7}-\frac{9}{2}log\mid x-4+\sqrt{x^2-8x+7}\mid+C

Explanation

Solution

Let I=∫x2−8x+7dxI=\int\sqrt{x^2-8x+7}dx

=∫(x2−8x+16)−9dx\int \sqrt{(x^2-8x+16)-9}dx

=∫(x−4)2−(3)2dx\int\sqrt{(x-4)^2-(3)^2}dx

It is known that,∫x2−a2dx=x2x2−a2−a22log⁡∣x+x2−a2∣+C\int\sqrt{x^2-a^2}dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log\mid x+\sqrt{x^2-a^2}\mid+C

∴I=(x−4)2x2−8x+7−92log⁡∣(x−4)+x2−8x+7∣+CI=\frac{(x-4)}{2}\sqrt{x^2-8x+7}-\frac{9}{2}\log\mid (x-4)+\sqrt{x^2-8x+7}\mid+C

Hence, the correct answer is D.