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  • Choose the correct answer in Exercises 10 to 11. Integral of root of x^2 - 8x + 7 dx is equal to

Choose the correct answer in Exercises 10 to 11.

    Q11.    \int \sqrt{x^2 - 8x+7}dx is equal to 

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

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

                (C)    \frac{1}{2}(x-4)\sqrt{x^2-8x+7} -3\sqrt2\log\left|x-4+\sqrt{x^2 -8x+7}\right| +C

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

Answers (1)

best_answer

Given integral  \int \sqrt{x^2 - 8x+7}dx

So, let us consider the function to be;

I = \int\sqrt{x^2-8x+7}dx =\int\sqrt{(x^2-8x+16)-(9)}dx

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

And 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

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

Therefore the correct answer is D.

Posted by

Divya Prakash Singh

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