how do you integrate
x^2/(9+x^6)
Let u = x^3, so that
du/3 = x^2 dx
The integral of x^2 dx/(9+x^6) then becomes
(1/3) Integral of du/(9+ u^2)
= (1/3)(1/3)tan^-1(u/3)
= (1/9)tan^-1(x^3/3)
x^2/(9+x^6)
du/3 = x^2 dx
The integral of x^2 dx/(9+x^6) then becomes
(1/3) Integral of du/(9+ u^2)
= (1/3)(1/3)tan^-1(u/3)
= (1/9)tan^-1(x^3/3)