<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] $$\int \frac{2-x^3}{(1+x^3)^{3/2}} dx=\int \frac{\frac{1}{x^3}(2-x^3)}{\frac1{x^3}\left(1+x^3\right)^{3/2}} \ dx$$
$$=\int \frac{\left(\frac{2}{x^3}-1\right)dx}{\left(\frac{1}{x^2}+x\right)^{3/2}}$$
$$=-\int \frac{d\left(\frac{1}{x^2}+x\right)}{\left(\frac{1}{x^2}+x\right)^{3/2}}$$
$$=- \frac{\left(\frac{1}{x^2}+x\right)^{-\frac32+1}}{-\frac32+1}+C$$
$$=\frac{2}{\sqrt{\frac1{x^2}+x}}+C$$
$$=\bbox[15px, #ffd, border:1px solid green]{\frac{2x}{\sqrt{1+x^3}}+C}$$
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