Question

Evaluate: $\underset{3}{\overset{5}{\int }}\frac{x^2}{x^2-4}dx$

• A. $2-{\log }_e\left(\frac{15}{7}\right)$
• B. $2+{\log }_e\left(\frac{15}{7}\right)$
• C. $2+4{\log }_{e}7+4{\log }_{e}5$
• D. $2-{\tan }^{-1}\left(\frac{15}{7}\right)$

Answer 1

The correct option is A $2-{\log }_e\left(\frac{15}{7}\right)$

$\frac{x^2}{x^2-4}=\frac{x^2-4+4}{x^2-4}=1+\frac{4}{x^2-4}=1+\frac{4}{(x+2)(x-2)}=1+\frac{1}{(x+2)}-\frac{1}{(x-2)}$

Thus, $\int \frac{x^2}{x^2-4}dx=\int (1+\frac{1}{(x+2)}-\frac{1}{(x-2)})dx$

${\int }_3^5\frac{x^2}{x^2-4}dx=[x+\log (x+2)-\log (x-2)]\stackrel{5}{\underset{3}{}}$

$=[5+\log 7-\log 3]-[3+\log 5-\log 1]=2+\log \frac{7}{15}=2-\log \frac{15}{7}$