Examples from Rationalising substitutions

$\int\frac{dx}{1+x^{1/4}}$

For the real principal fourth root, take $x\ge0$ and work on an interval with $x>0$ when differentiating the antiderivative.

Let $I=\int\frac{dx}{1+x^{1/4}}$. Set $u=x^{1/4}$, or $u^4=x$, so $dx=4u^3\, du$. Then

The sum of cubes is $1+u^3=(1+u)(1-u+u^2)$, hence

$\int\frac{dx}{\sqrt{e^{2x}-1}}$

For real values, the radicand requires $x\ge0$, but the denominator vanishes at $x=0$. Thus the integrand and its antiderivative are considered on $x>0$.

Let $I=\int\frac{dx}{\sqrt{e^{2x}-1}}$. Set $u=\sqrt{e^{2x}-1}>0$, so the denominator becomes $u$ and $e^{2x}=u^2+1$. Differentiating $u^2=e^{2x}-1$ gives $2u\,du=2e^{2x}\,dx$. Hence

$\int\frac{dx}{(1+e^{-x})^{3/2}}$

Let $I=\int\frac{dx}{(1+e^{-x})^{3/2}}$. Set $u^2=1+e^{-x}$. Differentiating gives $2u\, du=-e^{-x}\, dx$. Since $e^{-x}=u^2-1$, one has $-e^{-x}=1-u^2$, so

Here $u=\sqrt{1+e^{-x}}>1$ for every real $x$. Hence $(1+e^{-x})^{3/2}=(u^2)^{3/2}=u^3$ and $1-u^2=(1-u)(1+u)$. Thus

the minus sign coming from $1-u=-(u-1)$. Decompose

Clearing the denominator,

Setting $u=0$ gives $B=-1$. Setting $u=1$ gives $C=\frac12$. Setting $u=-1$ gives $D=-\frac12$. Setting $u=2$ gives $1=6A+3B+12C+4D$, hence $A=0$. Thus