∫displaystylex/(≤ft(x^2+3)≤ft(x^2+4)) dx =

The correct answer is Option (3) → \(\displaystyle\frac{1}{2}\log\left(\frac{x^2+3}{x^2+4}\right)+C\)

Let \(t = x^2\), so \(dt = 2x\,dx\):

    \( \displaystyle \int\frac{x}{\left(x^2+3\right)\left(x^2+4\right)}\,dx = \frac{1}{2}\int\frac{dt}{(t+3)(t+4)} \)

By partial fractions:

    \( \displaystyle \frac{1}{(t+3)(t+4)} = \frac{1}{t+3} - \frac{1}{t+4} \)

So

    \( \displaystyle \frac{1}{2}\int\left(\frac{1}{t+3} - \frac{1}{t+4}\right)dt = \frac{1}{2}\left(\log(t+3) - \log(t+4)\right) + C \)

    \( \displaystyle = \frac{1}{2}\log\left(\frac{x^2+3}{x^2+4}\right)+C \)

The moduli are not needed because \(x^2 + 3\) and \(x^2 + 4\) are always positive.