Integral Calculus Question 497
Question: The value of $ \int_{{}}^{{}}{\frac{dx}{\sqrt{x},(x+9)}dx} $ is equal to
[Pb. CET 2002]
Options:
A) $ {{\tan }^{-1}}\sqrt{x} $
B) $ {{\tan }^{-1}}( \frac{\sqrt{x}}{3} ) $
C) $ \frac{2}{3}{{\tan }^{-1}}\sqrt{x} $
D) $ \frac{2}{3}{{\tan }^{-1}}( \frac{\sqrt{x}}{3} ) $
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Answer:
Correct Answer: D
Solution:
We have, $ I=\int_{{}}^{{}}{\frac{dx}{\sqrt{x}(x+9)}} $ Put $ \sqrt{x}=t $ , squaring both sides, we get $ x=t^{2} $ and $ dx=2tdt $
$ \therefore $ $ I=2\int_{{}}^{{}}{\frac{dt}{t^{2}+3^{2}}}=\frac{2}{3}{{\tan }^{-1}}( \frac{t}{3} ) $
Þ $ I=\frac{2}{3}{{\tan }^{-1}}( \frac{\sqrt{x}}{3} ) $ .
BETA
