Evaluating $\int\frac{\sqrt[3]{x}}{1+x^2}\mathrm dx$

Question

Can anyone shed some light on evaluating the following integral? $$\int\frac{\sqrt[3]{x}}{1+x^2}\,\mathrm dx$$

Substitution $u = x^6$ does not get me any further or is it just my algebra?

Answer 1

Hint: with $x^2=u^3$ you get $2x\,\mbox{d}x=3u^2\,\mbox{d}u$ so $x^{1/3}\,\mbox{d}x=\tfrac{3}{2}u\,\mbox{d}u$ and the integral becomes: $$\int\frac{\sqrt[3]{x}}{1+x^2}\,\mbox{d}x=\frac{3}{2}\int\frac{u}{1+u^3}\,\mbox{d}u$$ Continuing with partial fractions may be a bit tedious, but it is pretty straightforward.

Answer 2

Another Hint: Take $$u=x^{1/3}$$ then, $$du= \frac{1}{3x^{2/3}}dx$$ hence the integral would be $$3\int\frac{u^3}{u^6+1}du$$

Answer 3

\begin{align} & u = \sqrt[3] x \\ & u^3 = x \\ & 3u^2\,du = dx \\ \\ & \int \frac{\sqrt[3]x}{1+x^2} \, dx = \int\frac u {1+u^6} 3u^2 \, du \end{align} $$ 1+u^6 = (u^2+1)(u^2-u\sqrt 3 + 1)(u^2+u\sqrt 3 + 1) $$ All three of these quadratic polynomials are irreducible unless one brings in complex numbers, as can be seen by completing the square or by looking at their discriminants.

Partial fractions should do the rest. I suspect there are also some really slick ways to do this even without resorting to thinking about residues, etc.