Definite Integration Question 254
Question: $ \int _{0}^{3}{\frac{3x+1}{x^{2}+9}dx=} $
[EAMCET 2003]
Options:
A) $ \log (2\sqrt{2})+\frac{\pi }{12} $
B) $ \log (2\sqrt{2})+\frac{\pi }{2} $
C) $ \log (2\sqrt{2})+\frac{\pi }{6} $
D) $ \log (2\sqrt{2})+\frac{\pi }{3} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_0^{3}{\frac{3x+1}{x^{2}+9}dx=\frac{3}{2}}\int_0^{3}{\frac{2x}{x^{2}+9}dx+}\int_0^{3}{\frac{dx}{x^{2}+9}} $
$ =[ \frac{3}{2}\log (x^{2}+9)+\frac{1}{3}{{\tan }^{-1}}( \frac{x}{3} ) ]_0^{3} $
$ =\frac{3}{2}(\log 18-\log 9)+\frac{1}{3}( \frac{\pi }{4} ) $
$ =\frac{3}{2}\log 2+\frac{\pi }{12}=\log (2\sqrt{2})+\frac{\pi }{12} $ .
BETA
