I'm stuck on this problem and have no idea on what to do.

The question states that where was to be a preliminary step before using partial fractions.

I have to integrate:


I've looked at the problem and have no idea on how to start it.


Make the substitution $u=e^x$. Then $u^2=e^{2x}$, $x=\ln u$ and $dx=\frac{du}{u}$. Your integral is then

$$\int \frac{dx}{e^x+e^{2x}}=\int\frac{du/u}{u+u^2}=\int \frac{du}{u^2+u^3}$$

Use partial fractions to continue from there.


$$\frac 1{e^x+e^{2x}} = \frac 1{e^x+(e^x)^2} = \frac 1{e^x(1+e^x)} = \frac A{e^x} + \frac B{1+e^x}$$ $$\implies 1=A(1+e^x) + B(e^x) \implies A=1 \wedge A+B=0$$


You want to find:

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

As you said, you want to find partial fractions to solve this. So to solve this:

$$\frac{1}{e^x+e^{2x}} = \frac{1}{e^x(1+e^{x})}=\frac{A}{e^x}+\frac{B}{1+e^x}$$





Which gives:


Since this equals 1, the $e^x$ terms cancel which means A=-B. This leaves to A=1 and in turn, B=-1

Thus we get the partial fraction


From there:

$$\int (\frac{1}{e^x}+\frac{-1}{1+e^x}) dx = \int \frac{1}{e^x}dx+\int\frac{-1}{1+e^x}dx $$

And this gives:

$$-e^{-x}+ln(e^x+1)-x + C$$

  • $\begingroup$ Hi, thanks for the input. I was just wondering on what you mean by " since this equals 1, the e^x cancel which means A=-B"? $\endgroup$ – Lagrange Feb 8 '15 at 1:45
  • $\begingroup$ For the equation $$A+Ae^x+Be^x=1$$ to work, since the right side has no '$e^x$' values, the '$e^x$' values on the left side must equal '0'. Thus $$Ae^x+Be^x=0$$ or $$A=-B$$ $\endgroup$ – Victor Feb 8 '15 at 1:47

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