\(\int \sqrt{x^2-8x+7}dx\) is equal to
\(\frac{1}{2}(x-4)\sqrt{x^2-8x+7}+9\log\mid x-4+\sqrt{x^2-8x+7}\mid+C\)
\(\frac{1}{2}(x+4)\sqrt{x^2-8x+7}+9\log\mid x+4+\sqrt{x^2-8x+7}\mid+C\)
\(\frac{1}{2}(x-4)\sqrt{x^2-8x+7}-3\sqrt 2\log\mid x-4+\sqrt{x^2-8x+7}\mid+C\)
\(\frac{1}{2}(x-4)\sqrt{x^2-8x+7}-\frac{9}{2}log\mid x-4+\sqrt{x^2-8x+7}\mid+C\)
Let \(I=\int\sqrt{x^2-8x+7}dx\)
=\(\int \sqrt{(x^2-8x+16)-9}dx\)
=\(\int\sqrt{(x-4)^2-(3)^2}dx\)
It is known that,\(\int\sqrt{x^2-a^2}dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log\mid x+\sqrt{x^2-a^2}\mid+C\)
∴\(I=\frac{(x-4)}{2}\sqrt{x^2-8x+7}-\frac{9}{2}\log\mid (x-4)+\sqrt{x^2-8x+7}\mid+C\)
Hence, the correct answer is D.
What is the Planning Process?
There are many important integration formulas which are applied to integrate many other standard integrals. In this article, we will take a look at the integrals of these particular functions and see how they are used in several other standard integrals.
These are tabulated below along with the meaning of each part.