What is the value of the definite integral of x^2/(x^3+1)^2 from x = 1 to x = 2?

Let f(x) = x^2/(x^3+1)^2

We
need to find the integral of f(x) from 1 to 2.

Let F(x) =
Int f(x)

==> The definite integral
is:

I = F(2) = F(1).

Let us
determine F(x).

==> F(x) = Int = x^2/(x^3+1)^2
dx

Let us assume that x^3+1 = u ==> du 3x^2
dx

==> F(x) = Int (1/u^2) *
du/3

               = Int du/
3u^2

                = (1/3) Int u^-2 du = (1/3) u^-1/-1 =
-1/3u

==> F(x) =
-1/3u

Now we will substitute back u= x^3
+1

==> F(x) =
-1/3(x^3+1)

==> F(2) = -1/3(8+1) =
-1/27

==> F(1) = -1/3(1+1) =
-1/6

==> I = -1/27 + 1/6 = -2/54 + 9/54 =
7/54

Then the definite integral is
7/54.