Definite integral of x^2/sqrt(x^3+1), between 2 and 3, gives=?

We'll apply Leibniz-Newton formula to evaluate the definite
integral:

Int f(x)dx = F(b) - F(a), where a and b are the limits of
integration.

First, we'll determine the indefinite integral. We'll
change the variable as method of solving the integral.

Int
x^2dx/sqrt(x^3 + 1)

We notice that if we'll put x^3 + 1 = t and
we'll differentiate, we'll get the numerator.

3x^2dx =
dt

x^2dx = dt/3

We'll re-write the
integral:

Int x^2dx/sqrt(x^3 + 1) = Int (dt/3)/sqrt
t

Int (dt/3)/sqrt t = (1/3)*Int dt/sqrt
t

(1/3)*Int dt/sqrt t = (1/3)*[t^(-1/2 + 1)/(-1/2 + 1)] +
C

(1/3)*Int dt/sqrt t = (2/3)*sqrt t +
C

Int f(x)dx = (2/3)*sqrt (x^3 + 1) +
C

We'll evaluate the definite integral, having as limits of
integration x = 2 and x = 3:

Int f(x)dx = (2/3)*sqrt (3^3 + 1) -
(2/3)*sqrt (2^3 + 1)

Int f(x)dx = (2/3)*(sqrt28 -
3)

Int f(x)dx = (2*sqrt28)/3 -
2

The definite integral of the function f(x) =
x^2/sqrt(x^3 + 1), is Int f(x)dx = (2*sqrt28)/3 - 2.