simple integral

1.

Again it's getting lengthy. A very complicated use of partial fraction has to be made I guess.

I could complete it till here:

I = ∫dx(1+x)(1-x+x²) -∫dx(1+x)²(1-x+x²)²
I₁ I₂

I could evaluate I₁ by Partial fractions. But Didn't try for I₂.

Anyone to complete this or via other method.

\hspace{-16}\int\frac{x^3}{(1+x^3)^2}dx$\\\\\\ $\frac{1}{3}.\int x.\frac{3x^2}{(1+x^3)^2}dx$\\\\ Now Using Integration by parts, We Get\\\\ $-\frac{1}{3}.x.\frac{1}{(1+x^3)}+\frac{1}{3}.\int\frac{1}{1+x^3}dx$\\\\

Very Nice.