Q. The value of $\int_{3}^{5}\frac{x^{2}}{x^{2} - 4}\ dx$ is

A) $2 - \log_{e}\left( \frac{15}{7} \right)$
B) $2 + \log_{e}\left( \frac{15}{7} \right)$
C) $2 + 4\log_{e}{3 - 4\log_{e}{7 + 4\log_{e}5}}$
D) $2 - \tan^{- 1}\left( \frac{15}{7} \right)$
Explanation

(b)
Let $I = \int_{3}^{5}\frac{x^{2}}{x^{2} - 4}\ dx = \int_{3}^{5}\left( \frac{x^{2} - 4}{x^{2} - 4} + \frac{4}{x^{2} - 4} \right)dx$
$= \int_{3}^{5}\left( 1 + \frac{4}{x^{2} - 4} \right)dx$
$= \left\lbrack x + \frac{4}{2 \times 2}\log_{e}\left( \frac{x - 2}{x + 2} \right) \right\rbrack_{3}^{5}$
$= \left\lbrack 5 + \log_{e}\left( \frac{5 - 2}{5 + 2} \right) - 3 - \log_{e}\left( \frac{3 - 2}{3 + 2} \right) \right\rbrack$
$= 2 + \log_{e}\left( \frac{3}{7} \right) - \log_{e}\left( \frac{1}{5} \right)$
$= 2 + \log_{e}\left( \frac{3}{7} \times \frac{5}{1} \right) = 2 + \log_{e}\left( \frac{15}{7} \right)$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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