The value of $\int\limits_2^4\frac{x}{x^2+1}dx$ is equal to

$\frac{1}{2}\log_e(\frac{17}{5})$

The correct answer is Option (1) → $\frac{1}{2}\log_e(\frac{17}{5})$ **

Compute $\displaystyle \int_{2}^{4}\frac{x}{x^{2}+1}\,dx$

Use substitution: let $u=x^{2}+1$

$du=2x\,dx$

$\frac{x\,dx}{x^{2}+1}=\frac{1}{2}\frac{du}{u}$

When $x=2$, $u=5$

When $x=4$, $u=17$

$\displaystyle \int_{2}^{4}\frac{x}{x^{2}+1}\,dx=\frac{1}{2}\int_{5}^{17}\frac{du}{u}$

$=\frac{1}{2}\left[\log_e u\right]_{5}^{17}$

$=\frac{1}{2}\log_e\left(\frac{17}{5}\right)$

$\frac{1}{2}\log_e\left(\frac{17}{5}\right)$