The value of the definite integral I =∫limits_1^21/(x(1+x^2))dx is:

The correct answer is Option (2) → $\frac{3}{2}\log 2-\frac{1}{2}\log 5$

$I=\int_{1}^{2}\frac{1}{x(1+x^{2})}\,dx$

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

$I=\int_{1}^{2}\left(\frac{1}{x}-\frac{x}{1+x^{2}}\right)\,dx$

$=\left[\ln x-\frac{1}{2}\ln(1+x^{2})\right]_{1}^{2}$

$=\ln2-\frac{1}{2}\ln5-\left(\ln1-\frac{1}{2}\ln2\right)$

$=\ln2-\frac{1}{2}\ln5+\frac{1}{2}\ln2$

$=\frac{3}{2}\ln2-\frac{1}{2}\ln5$

$=\frac{1}{2}\ln\!\left(\frac{8}{5}\right)$

The value of the integral is $\frac{1}{2}\ln\left(\frac{8}{5}\right) = \frac{3}{2}\log 2-\frac{1}{2}\log 5$.