Ratio of two polynomial functions - integral

[Perpendicular]

The problem is :-

Integral of (1+x^4) / ( 1 + x^6) . dx

I have reduced it to a form of

Integral of 2.sqrt(tantheta) / ( 1 + tan^3 theta ) over dtheta where x^2 = tantheta.

However I cannot reduce it further. How do I proceed ? In general, how do I proceed given a problem of the form of finding the integral of two polynomials when neither is a simple quadratic or factorized ?

 

[DonAntonio]

The problem is :-

Integral of (1+x^4) / ( 1 + x^6) . dx

I have reduced it to a form of

Integral of 2.sqrt(tantheta) / ( 1 + tan^3 theta ) over dtheta where x^2 = tantheta.

However I cannot reduce it further. How do I proceed ? In general, how do I proceed given a problem of the form of finding the integral of two polynomials when neither is a simple quadratic or factorized ?

If you first find all the solutions to the complex equation $\,z^6+1=0\,$ (i.e., the 6 roots of -1), then you find them to be $$z_1=i\,,\,\overline{z_1}=-i\,,\,z_2=\frac{1}{2}(\sqrt{3}+i)\,,\,\overline{z_2}=\frac{1}{2}(\sqrt{3}-i)\,,\,z_3=-\frac{1}{2}(\sqrt{3}-i)\,,\,\overline{z_3}=-\frac{1}{2}(\sqrt{3}+i)$$ As you can see, the roots come in conjugate paris, so from here you can factorise the polynomial over the reals: $$x^6+1=(x-z_1)(x-\overline{z_1})(x-z_2)(x-\overline{z_2})(x-z_3)(z-\overline{z_3})=(x^2+1)(x^2-\sqrt{3}\,x+1)(x^2+\sqrt{3}\,x+1)$$and now you can do partial fractions:$$\frac{x^4+1}{x^6+1+}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2-\sqrt{3}\,x+1}+\frac{Ex+F}{x^2+\sqrt{3}\,x+1}$$

DonAntonio