\(\int \sqrt{x^2-8x+7}dx\) is equal to
\(\int \sqrt{x^2-8x+7}dx\) is equal to
\(\frac{1}{2}(x-4)\sqrt{x^2-8x+7}+9\log\mid x-4+\sqrt{x^2-8x+7}\mid+C\)
\(\frac{1}{2}(x+4)\sqrt{x^2-8x+7}+9\log\mid x+4+\sqrt{x^2-8x+7}\mid+C\)
\(\frac{1}{2}(x-4)\sqrt{x^2-8x+7}-3\sqrt 2\log\mid x-4+\sqrt{x^2-8x+7}\mid+C\)
\(\frac{1}{2}(x-4)\sqrt{x^2-8x+7}-\frac{9}{2}log\mid x-4+\sqrt{x^2-8x+7}\mid+C\)
The Correct Option is D
Solution and Explanation
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.
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Show that the relation R in the set R of real numbers, defined as
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Concepts Used:
Integrals of Some Particular Functions
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.
Integrals of Some Particular Functions:
- ∫1/(x2 – a2) dx = (1/2a) log|(x – a)/(x + a)| + C
- ∫1/(a2 – x2) dx = (1/2a) log|(a + x)/(a – x)| + C
- ∫1/(x2 + a2) dx = (1/a) tan-1(x/a) + C
- ∫1/√(x2 – a2) dx = log|x + √(x2 – a2)| + C
- ∫1/√(a2 – x2) dx = sin-1(x/a) + C
- ∫1/√(x2 + a2) dx = log|x + √(x2 + a2)| + C
These are tabulated below along with the meaning of each part.
