The value of the definite integral \(I =\int_{1}^{2}\displaystyle\frac{1}{x(1+x^2)}dx\) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{3}{2}\log 2-\frac{1}{2}\log 5\)
By partial fractions, since \((1 + x^2) - x^2 = 1\),
\( \displaystyle \frac{1}{x(1+x^2)} = \frac{(1 + x^2) - x^2}{x(1+x^2)} = \frac{1}{x} - \frac{x}{1+x^2} \)
\( \displaystyle I = \int_{1}^{2} \left(\frac{1}{x} - \frac{x}{1+x^2}\right) dx = \left[\log x - \frac{1}{2}\log(1+x^2)\right]_{1}^{2} \)
\( \displaystyle = \left(\log 2 - \frac{1}{2}\log 5\right) - \left(0 - \frac{1}{2}\log 2\right) \)
\( \displaystyle = \frac{3}{2}\log 2 - \frac{1}{2}\log 5 \)