Question

The value of ${\int }_3^5\frac{x^2}{x^2-4}dx=?$

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

Answer 1

The correct option is B

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

Explanation for correct option:

Compute the required value:

Given: ${\int }_3^5\frac{x^2}{x^2-4}dx$

$$\begin{gathered} \Rightarrow {\int }_3^5\frac{x^2-4+4}{x^2-4}dx \\ \Rightarrow {\int }_3^5\frac{x^2-4}{x^2-4}dx+{\int }_3^5\frac{4}{x^2-4}dx \\ \Rightarrow {\int }_3^{5}1·dx+\frac{4}{4}{\left|\log \left(\frac{x-2}{x+2}\right)\right|}_3^5;\left[\int \frac{1}{x^2-a^2}·dx=\frac{1}{2a}\log \left(\frac{x-a}{x+a}\right)\right] \\ \Rightarrow {\left|x\right|}_3^5+\left|\log \left(\frac{5-2}{5+2}\right)-\log \left(\frac{3-2}{3+2}\right)\right| \\ \Rightarrow 5-3+\log \left(\frac{3}{7}\right)-\log \left(\frac{1}{5}\right) \\ \Rightarrow 2+\log \left(\frac{3\times 5}{7}\right);\left[\log \left(m\right)-\log \left(n\right)=\log \left(\frac{m}{n}\right)\right] \\ \Rightarrow 2+\log \left(\frac{15}{7}\right) \end{gathered}$$

Hence, option $B$ is the correct answer.