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) \[\frac{1}{5}\log \frac{3}{5}\]
Correct Answer: A
Solution :
\[I=\int_{8}^{15}{\frac{dx}{(x-3)\sqrt{x+1}}}\] Put \[x={{\tan }^{2}}\theta \Rightarrow \theta ={{\tan }^{-1}}\sqrt{x}\] \[dx=2\tan \theta {{\sec }^{2}}\theta \,d\theta \] \[\therefore \]\[I=\int_{{{\tan }^{-1}}\sqrt{8}}^{{{\tan }^{-1}}\sqrt{15}}{\frac{2\tan \theta {{\sec }^{2}}\theta \,}{({{\tan }^{2}}\theta -3)\sqrt{{{\tan }^{2}}\theta +1}}d\theta }\] \[=\int_{{{\tan }^{-1}}\sqrt{8}}^{{{\tan }^{-1}}\sqrt{15}}{\frac{2\tan \theta {{\sec }^{2}}\theta \,}{({{\sec }^{2}}\theta -4)\sec \theta }d\theta }\] \[=\int_{{{\tan }^{-1}}\sqrt{8}}^{{{\tan }^{-1}}\sqrt{15}}{\frac{2\tan \theta \sec \theta }{({{\sec }^{2}}\theta -4)}\,d\theta }\] \[=\int_{{{\tan }^{-1}}\sqrt{8}}^{{{\tan }^{-1}}\sqrt{15}}{\frac{2\tan \theta \sec \theta }{(\sec \theta -2)(\sec \theta +2)}\,d\theta }\] \[=\left[ \frac{1}{2}\log \frac{(\sec \theta -2)}{(\sec \theta +2)} \right]_{{{\tan }^{-1}}\sqrt{8}}^{{{\tan }^{-1}}\sqrt{15}}\] \[=\frac{1}{2}\left[ \log \frac{2}{6}-\log \frac{1}{5} \right]=\frac{1}{2}\log \frac{5}{3}\].
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