The value of the integral $\int\limits_2^4 \frac{x}{x^2+1} dx$ is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) - $\frac{1}{2} \log \left(\frac{17}{5}\right)$
$I=\frac{1}{2}\int\limits_2^4 \frac{2x}{x^2+1} dx$
$I=\frac{1}{2}\left[\log|x^2+1|\right]_2^4=\frac{1}{2}\log|\frac{17}{5}|$