The value of $\int_1^3 \frac{x^2}{x^3+1}dx$

Look at the integral: $\int_1^3 \frac{x^2}{x^3+1}dx$

The denominator is $x^3 + 1$ and the numerator is $x^2$. The derivative of $x^3 + 1$ is $3x^2$, which is almost what appears in the numerator.

Let $u = x^3 + 1$

$\frac{du}{dx} = 3x^2$

$du = 3x^2 \, dx$

$x^2 \, dx = \frac{1}{3} du$

The limits change from $x$ to $u$:

When $x = 1$: $u = (1)^3 + 1 = 2$

When $x = 3$: $u = (3)^3 + 1 = 28$

$\int_1^3 \frac{x^2}{x^3+1}dx = \int_2^{28} \frac{1}{u} \cdot \frac{1}{3} du$

$= \frac{1}{3} \int_2^{28} \frac{1}{u} \, du$

$= \frac{1}{3} [\log u]_2^{28}$

$= \frac{1}{3} (\log 28 - \log 2)$

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

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

Therefore, $\int_1^3 \frac{x^2}{x^3+1}dx = \frac{1}{3}\log 14$