Currently I'm facing a lot of integrals of the form $$I(p)=\int\frac{dx}{p\left(e^x\right)}$$ where $p:\mathbb{R}\to\mathbb{R}$ is a polynomial. For example, $$\begin{split}I(x+1)&=\int\frac{dx}{e^x+1} \\&=\int\frac{e^{-x}dx}{1+e^{-x}}\\&=\ln\left(1+e^{-x}\right)+C\end{split}$$ but for $p(x)=x^2+x+1$ the problem is already getting tricky. I'm trying to find a general method to this problem. I tried to make the Ansatz (where $q$ need not be a polynomial - it can be anything!) $$I(p)=\ln \left(q\left(e^x\right)\right)$$ which leads to (taking derivatives on both sides) $$\frac{1}{p\left(e^x\right)}=\frac{q'\left(e^x\right)e^x}{q\left(e^x\right)}$$ which is (almost) the same as the problem $$\frac{q'(x)}{q(x)}=\frac{1}{xp(x)}$$ which has a 'solution' (integrating both sides and taking exponentials) $$q(x)=\exp\left(\int\frac{dx}{xp(x)}\right),$$ but studying rational integrands hasn't really a general approach so it doesn't seem that this method is getting me anywhere! I hope another Ansatz - or maybe a completely different approach - exists to deal with this problem.


We have $$\int\frac{dx}{p(e^x)}=\int\frac{e^x\,dx}{e^xp(e^x)}.$$ Then substituting $t=e^x$ we arrive at $$\int\frac{dt}{q(t)},$$where $q(t)=tp(t)$ is also a polynomial. Then use the standard methods: partial fractions and so on.

  • $\begingroup$ Thanks! Do you think that it is likely that a better approach exists? Because if it were possible then the problem of integrating rational functions would suddenly be more efficient as well... $\endgroup$ – RMWGNE96 May 12 at 19:45
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    $\begingroup$ @RMWGNE96 I think partial fractions is efficient enough for rational functions. $\endgroup$ – Julian Mejia May 12 at 19:48
  • $\begingroup$ @JulianMejia true $\endgroup$ – RMWGNE96 May 12 at 19:49

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