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In my textbook, there is the following integral: $\int \frac{x^3}{1+x^2}dx$, the next line is:$\frac{1}{2}\int \frac{x^2}{1+x^2}dx^2$, my question is what exactly did the author do? Which theorem did he used? How did he increased the degree in the differential(how from $dx$, he got to $dx^2$)

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    $\begingroup$ $d(x^2)=2x\,dx$. $\endgroup$ – Gerry Myerson Feb 6 '17 at 9:51
  • $\begingroup$ $dx^2/dx=2x$, so $dx^2=2xdx$, and you can substitute $x^3 dx$ into $\frac{1}{2} x^2 dx^2$. $\endgroup$ – user156937 Feb 6 '17 at 9:51
  • $\begingroup$ They used $x\, dx = \frac{1}{2}\, dx^2$ because $\frac{dx^2}{dx}=2x$ (differentiate $x^2$ with respect to $x$). There shouldn't be a $-$ in $-\frac{1}{2}$. $\endgroup$ – user236182 Feb 6 '17 at 9:52
  • $\begingroup$ See also: math.stackexchange.com/questions/960298/… $\endgroup$ – lab bhattacharjee Feb 6 '17 at 9:53
  • $\begingroup$ One may see this as a change of variable: $u=x^2$, $du=2xdx$. $\endgroup$ – Olivier Oloa Feb 6 '17 at 9:54
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It is just a (often misleading) way to perform the substitution $x=\sqrt{u}$, $dx=\frac{du}{2\sqrt{u}}$ $$ \int \frac{x^3}{1+x^2}\,dx = \frac{1}{2}\int \frac{u^{3/2}}{(1+u)\,u^{1/2}}\,du=\frac{1}{2}\int\frac{u}{1+u}\,du.$$ This clearly leads to: $$ \int \frac{x^3}{1+x^2}\,dx = C+\frac{x^2-\log(1+x^2)}{2}.$$

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