1. ## Integral

$\displaystyle \int\frac{dx}{e^x+e^{2x}}$
$\displaystyle e^x=t$
$\displaystyle dx=\frac{dt}{t}$
$\displaystyle \int\frac{dt}{t^2(1-t)}$
What's next ?
Is using partial fractions a good idea ?

2. Hello, totalnewbie!

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

$\displaystyle e^x = t\quad\Rightarrow\quad dx = \frac{dt}{t}\quad\Rightarrow\quad \int\frac{dt}{t^2(1-t)}$ . . . oops! should be 1 + t

What's next? . Is using partial fractions a good idea?

Yes . . . an excellent idea!

3. then
$\displaystyle \frac{A}{t^2}+\frac{B}{1-t}=\frac{1}{t^2*(1-t)}$
$\displaystyle A=1$
$\displaystyle B=1$

then I got:
$\displaystyle \int\frac{dx}{e^x+e^{2x}}=\int t^{-2}+ln(1-t)=-e^{-x}+ln(1-e^x)+ C$

4. Originally Posted by totalnewbie
then
$\displaystyle \frac{A}{t^2}+\frac{B}{1-t}=\frac{1}{t^2*(1-t)}$
$\displaystyle A=1$
$\displaystyle B=1$

then I got:
$\displaystyle \int\frac{dx}{e^x+e^{2x}}=\int t^{-2}+ln(1-t)=-e^{-x}+ln(1-e^x)+ C$

Why is there a minus, maybe it should be $\displaystyle 1+t$

5. Originally Posted by ThePerfectHacker
Why is there a minus, maybe it should be $\displaystyle 1+t$
oops, then it's
$\displaystyle \int\frac{dx}{e^x+e^{2x}}=\int t^{-2}+ln(1+t)=-e^{-x}+ln(1+e^x)+ C$
but despite it doesn't match with function calculator answer.

6. Hello again, totalnewbie!

You're missing the "repeated factor" in your Partial Fractions.

You should have had: .$\displaystyle \frac{1}{t^2(1 + t)} \;=\;\frac{A}{t} + \frac{B}{t^2} + \frac{C}{1+t}$

. . and get: .$\displaystyle A = -1,\;B = 1,\;C = 1$

Then integrate: .$\displaystyle \int\left(-\frac{1}{t} + \frac{1}{t^2} + \frac{1}{1+t}\right)dt \;=\;-\int\frac{dt}{t} + \int t^{-2}dt + \int\frac{dt}{1+t}$