Question

Solve $\int \frac{x^3-x}{1+x^6}dx$ is

Answer 1

$\int \frac{x^3-x}{1+x^6}dx=\int \frac{1-\frac{1}{x^2}}{1+\frac{1}{x^6}}\frac{dx}{x^3}$

Put $t=\frac{1}{x^2}⟹dt=-2\frac{dx}{x^3}$

$\int \frac{1-t}{1+t^3}(-\frac{dt}{2})=-\frac{1}{3}\int \frac{1}{1+t}dt-\frac{1}{3}\int \frac{2t+1}{t^2-t+1}dt=-\frac{1}{3}\ln (1+t)-\frac{1}{3}\int \frac{2(t-\frac{1}{2})}{(t-\frac{1}{2}{)}^2+\frac{3}{4}}dt-\frac{2}{3}\int \frac{dt}{(t-\frac{1}{2}{)}^2+\frac{3}{4}}$

$=-\frac{1}{3}\ln (1+t)-\frac{1}{3}\ln (t^2-t+1)-\frac{4}{3\sqrt{3}}{\text{tan}}^{-1}(\frac{2t-1}{\sqrt{3}})+C$