calculus - Show that $int^infty_1frac{1}{1+x^3}dx = int^1_0frac{x}{1+x^3}dx$

Consider the integral $$J = \int^\infty_0\frac{1}{1+x^3}dx$$ show that $$\int^\infty_1\frac{1}{1+x^3}dx = \int^1_0\frac{x}{1+x^3}dx$$ and then deduce that $$J = \int^1_0f(x) dx$$ where f is a function to be determined.

I'm specifically stuck on the second part of the question. It is easy to miss but the bounds for J is $0$ and $\infty$ and not $1$ and $\infty$ as in the case of the first part of the question.

Answer

Let ,

$x = \frac{1}{t}$ , $dx = \frac{-1}{t^2}dt$

at $x = \infty, t = 0$

at $x = 1, t= 1$

$I = \int^0_1\frac{1}{1+\frac{1}{t^3}}.\frac{-dt}{t^2} = - \int^0_1\frac{t^3dt}{(t^3 + 1)t^2}$

Changing the limits,

$I = \int^1_0 \frac{tdt}{1+t^3} = \int^1_0\frac{xdx}{1+x^3}$ (Replacing t by x)

$J = \int^\infty_0\frac{dx}{1+x^3} = \int^1_0\frac{dx}{1+x^3}+ \int^\infty_1\frac{dx}{1+x^3} = \int^1_0\frac{dx}{1+x^3} + \int^1_0\frac{xdx}{1+x^3}$ (From I)

$J = \int^1_0\frac{(x+1)dx}{1+x^3}$

Thus, $f(x) = \frac{x+1}{1+x^3}$