Definite Integration Question 204
Question: $ \int _{8}^{15}{\frac{dx}{(x-3)\sqrt{x+1}}=} $
[UPSEAT 2003]
Options:
A) $ \frac{1}{2}\log \frac{5}{3} $
B) $ \frac{1}{3}\log \frac{5}{3} $
C) $ \frac{1}{2}\log \frac{3}{5} $
D) $ \frac{1}{5}\log \frac{3}{5} $
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Answer:
Correct Answer: A
Solution:
$ I=\int_8^{15}{\frac{dx}{(x-3)\sqrt{x+1}}} $
Put $ x={{\tan }^{2}}\theta \Rightarrow \theta ={{\tan }^{-1}}\sqrt{x} $
$ dx=2\tan \theta {{\sec }^{2}}\theta d\theta $
$ \therefore $ $ I=\int _{{{\tan }^{-1}}\sqrt{8}}^{{{\tan }^{-1}}\sqrt{15}}{\frac{2\tan \theta {{\sec }^{2}}\theta }{({{\tan }^{2}}\theta -3)\sqrt{{{\tan }^{2}}\theta +1}}d\theta } $
$ =\int _{{{\tan }^{-1}}\sqrt{8}}^{{{\tan }^{-1}}\sqrt{15}}{\frac{2\tan \theta {{\sec }^{2}}\theta }{({{\sec }^{2}}\theta -4)\sec \theta }d\theta } $
$ =\int _{{{\tan }^{-1}}\sqrt{8}}^{{{\tan }^{-1}}\sqrt{15}}{\frac{2\tan \theta \sec \theta }{({{\sec }^{2}}\theta -4)}d\theta } $
$ =\int _{{{\tan }^{-1}}\sqrt{8}}^{{{\tan }^{-1}}\sqrt{15}}{\frac{2\tan \theta \sec \theta }{(\sec \theta -2)(\sec \theta +2)}d\theta } $
$ =[ \frac{1}{2}\log \frac{(\sec \theta -2)}{(\sec \theta +2)} ] _{{{\tan }^{-1}}\sqrt{8}}^{{{\tan }^{-1}}\sqrt{15}} $
$ =\frac{1}{2}[ \log \frac{2}{6}-\log \frac{1}{5} ]=\frac{1}{2}\log \frac{5}{3} $ .
BETA
