<s>[INST] Evaluate $\int \frac{2-x^3}{(1+x^3)^{3/2}} dx$ 
Evaluate:
$$\int \frac{2-x^3}{(1+x^3)^{3/2}} dx$$

I could find the integral by setting it equal to $$\frac{ax+b}{(1+x^3)^{1/2}}$$
and differentiating both sides w.r.t.$x$ as
$$\frac{2-x^3}{(1+x^3)^{3/2}}=\frac{a(1+x^3)^{3/2}-(1/2)(ax+b)3x^2(1+x^3)^{-1/2}}{(1+x^3)}$$$$=\frac{a-ax^3/2-3bx^2}{(1+x^3)^{3/2}}$$
Finally by setting $a=2,b=0$, we get $$I(x)=\frac{2x}{(1+x^3)^{1/2}}+C$$
The question is: How to do it otherswise?
 [/INST] Split the integral and integrate by parts:
$$\begin{align}I = \int\dfrac{2 - x^3}{\left(1 + x^3\right)^{3/2}}\,\mathrm dx &\equiv\int\dfrac2{\left(1 + x^3\right)^{3/2}}\,\mathrm dx - \int\dfrac{x^2}{\left(1 + x^3\right)^{3/2}}x\,\mathrm dx \\ &= \int\dfrac2{\left(1 + x^3\right)^{3/2}}\,\mathrm dx - \left(-\dfrac{2x}{3\left(1 + x^3\right)^{1/2}} + \dfrac23\int\dfrac{1 + x^3}{\left(1 + x^3\right)^{3/2}}\,\mathrm dx\right) \\ &= \int\dfrac{6 - 2 - 2x^3}{3\left(1 + x^3\right)^{3/2}}\,\mathrm dx + \dfrac{2x}{3\sqrt{1 + x^3}} \\ &= \dfrac23\int\dfrac{2 - x^3}{\left(1 + x^3\right)^{3/2}}\,\mathrm dx + \dfrac{2x}{3\sqrt{1 + x^3}} \\ &= \dfrac23I + \dfrac{2x}{3\sqrt{1 + x^3}} \\ \implies I &= \dfrac{2x}{\sqrt{1 + x^3}}\end{align}$$
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