Integration

hint : divide numerator and denominator by x²

\hspace{-16}\bf{\int\frac{x^3-2}{\sqrt{(x^3+1)^3}}dx}$\\\\\\ Using Hit and Trail......\\\\ Let $\bf{\frac{d}{dx}\left\{\frac{Ax+B}{\sqrt{x^3+1}}\right\}=\frac{x^3-2}{\sqrt{(x^3+1)^3}}}$\\\\\\ $\bf{\frac{2(x^3+1).A-(Ax+B).3x^2}{\sqrt{(x^3+1)^3}}=\frac{2x^3-4}{\sqrt{(x^3+1)^3}}}$\\\\\\ $\bf{\frac{-Ax^3+2A-3Bx^2}{\sqrt{(x^3+1)^3}}=\frac{2x^3-4}{\sqrt{(x^3+1)^3}}}$\\\\\\ Now Camparing, We Get\\\\ $\bf{A=-2\;\;,B=0}$\\\\\\ So $\bf{\frac{d}{dx}\left\{\frac{-x+0}{\sqrt{x^3+1}}\right\}=\frac{x^3-2}{\sqrt{(x^3+1)^3}}}$\\\\\\ Now Integrate Both Side, We Get\\\\ $\bf{\int \frac{d}{dx}\left\{\frac{-2x}{\sqrt{x^3+1}}\right\}dx=\int \frac{x^3-2}{\sqrt{(x^3+1)^3}}dx}$\\\\\\ So $\bf{\int \frac{x^3-2}{\sqrt{(x^3+1)^3}}dx=\frac{-2x}{\sqrt{x^3+1}}+C}$

ignore my above post ,

dividing by x³ and taking x+1/x² =z also does the trick.

It is dividing by x³ not x²