2
$\begingroup$

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 $dx^2$)

  • 2
    $d(x^2)=2x\,dx$. – 2017-02-06
  • 0
    $dx^2/dx=2x$, so $dx^2=2xdx$, and you can substitute $x^3 dx$ into $\frac{1}{2} x^2 dx^2$. – 2017-02-06
  • 0
    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 $-\frac{1}{2}$. – 2017-02-06
  • 0
    See also: http://math.stackexchange.com/questions/960298/integrate-int-fracx3dx-sqrt1x2 – 2017-02-06
  • 0
    One may see this as a change of variable: $u=x^2$, $du=2xdx$. – 2017-02-06
  • 0
    It could be clearer writing $\frac{1}{2}\int \frac{x^2}{1+x^2}\,d(x^2)$ – 2017-02-06
  • 0
    Because the author would like to do the following:$$\int\frac{x^3}{1+x^2}dx=\int\frac{x^2x}{1+x^2}dx=\frac{1}{2}\int\frac{x^2\cdot 2xdx}{1+x^2}.$$ Did you notice $2xdx$? This is exactly the differential of $x^2$, in symbol, we write $$2xdx=d(x^2).$$ The author wrote $dx^2$ to mean $d(x^2)$. – 2017-02-06

1 Answers 1

2

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: $$ \int \frac{x^3}{1+x^2}\,dx = C+\frac{x^2-\log(1+x^2)}{2}.$$