\(\int \sqrt{x^2-8x+7}dx\) 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.
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]


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.
