The value of ∫_1^3(x^2)/(x^3+1) dx

The correct answer is Option (3) → \(\displaystyle\frac{1}{3}\log 14\)

The numerator is a constant multiple of the derivative of the denominator, so let \(u = x^3 + 1\):

    \(du = 3x^2\,dx\)

    \(x = 1 \Rightarrow u = 2,\quad x = 3 \Rightarrow u = 28\)

So:

    \( \displaystyle \int_{1}^{3}\frac{x^2}{x^3+1}\,dx = \frac{1}{3}\int_{2}^{28}\frac{du}{u} \)

    \( \displaystyle = \frac{1}{3}\left[\log u\right]_{2}^{28} \)

    \( \displaystyle = \frac{1}{3}\left(\log 28 - \log 2\right) = \frac{1}{3}\log\frac{28}{2} = \frac{1}{3}\log 14 \)