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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Choose the correct answer in $ \Large \int \normalsize \sqrt{x^2-8x+7} \: dx$ is equal to\[ \begin{array}{l} (A)\frac{1}{2}(x-4)\sqrt{x^2-8x+7}+9\;log\mid x-4+\sqrt{x^2-8x+7}\mid + \; C \\ (B)\frac{1}{2}(x+4)\sqrt{x^2-8x+7}+9\;log\mid x+4+\sqrt{x^2-8x+7}\mid + \; C \\ (C) \frac{1}{2}(x-4)\sqrt{x^2-8x+7}-3\sqrt2log\mid x-4+\sqrt{x^2-8x+7}\mid+ \; C \\ (D)\frac{1}{2}(x-4)\sqrt{x^2-8x+7}-\frac{9}{2}\;log\mid x-4+\sqrt{x^2-8x+7}\mid + \; C \end{array}\]

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1 Answer

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  • $\int \sqrt{x^2-a^2}dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac {a^2}{2}log |x+\sqrt{x^2-a^2}|+c$
Given $I=\int \sqrt{x^2-8x+7}dx$
 
Here we can split $ x^2-8x+7$ as $(x^2-8x+16)-16+7$
 
$=(x-4)^2-9$
 
$=(x-4)^2-3^2$
 
Therefore $I=\int [(x-4)^2-3^2]dx$
 
Clearly this is of the form $\sqrt{x^2-a^2}=$
 
$=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}log|x+\sqrt{x^2-a^2}|+c$
 
Here in the place of x we have x-4 and in the place of a we have 3
 
Hence $\int\sqrt{x^2-8x+7} dx=\frac{(x-4)}{2}\sqrt{x^2-8x+7}-\frac{9}{2} log |(x-4)+\sqrt{x^2-8x+7}|+c$
 
Hence the correct answer is D

 

answered Feb 8, 2013 by meena.p
 

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