The value of $\int\limits_1^3\frac{x^2}{x^3+1}dx$

$\frac{1}{3}\log 14$

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

$\int_{1}^{3} \frac{x^2}{x^3 + 1} \, \mathrm{d}x$

Let $u = x^3 + 1$.

$\frac{d}{dx}(x^3 + 1) = 3x^2 \Rightarrow \frac{du}{3} = x^2 \, \mathrm{d}x$.

When $x = 1$, $u = 2$.

When $x = 3$, $u = 28$.

Substitute in the integral:

$= \int_{2}^{28} \frac{1}{u} \frac{du}{3}$

$= \frac{1}{3} \int_{2}^{28} \frac{1}{u} \, \mathrm{d}u$

$= \frac{1}{3} \ln |u| \bigg|_{2}^{28}$

$= \frac{1}{3} \left( \ln 28 - \ln 2 \right)$

$= \frac{1}{3} \ln \frac{28}{2}$

$= \frac{1}{3} \ln 14$