The value of int_3^5 {frac{{{x^2}}}{{{x^2} - | vedclass

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(B) Given integral $I = \int_3^5 \frac{x^2}{x^2 - 4} \, dx$. We can rewrite the integrand as: $\frac{x^2}{x^2 - 4} = \frac{x^2 - 4 + 4}{x^2 - 4} = 1 +

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$2 + \log_e \left( \frac{15}{7} \right)$

Vedclass · Answer

(B) Given integral $I = \int_3^5 \frac{x^2}{x^2 - 4} \, dx$. We can rewrite the integrand as: $\frac{x^2}{x^2 - 4} = \frac{x^2 - 4 + 4}{x^2 - 4} = 1 + \frac{4}{x^2 - 4}$. Now,$I = \int_3^5 \left( 1 + \frac{4}{x^2 - 4} \right) \, dx$. Using the partial fraction decomposition $\frac{4}{x^2 - 4} = \frac{4}{(x-2)(x+2)} = \frac{1}{x-2} - \frac{1}{x+2}$. So,$I = \int_3^5 \left( 1 + \frac{1}{x-2} - \frac{1}{x+2} \right) \, dx$. Integrating term by term: $I = [x + \log_e |x-2| - \log_e |x+2|]_3^5$. $I = [x + \log_e \left| \frac{x-2}{x+2} \right|]_3^5$. Evaluating at the limits: $I = (5 + \log_e \frac{3}{7}) - (3 + \log_e \frac{1}{5})$. $I = 2 + \log_e \left( \frac{3/7}{1/5} \right) = 2 + \log_e \left( \frac{15}{7} \right)$. Thus,the correct option is $B$.

The value of $\int_3^5 {\frac{{{x^2}}}{{{x^2} - 4}}\,dx} $ is

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