$\displaystyle \int\frac{e^{2x}}{e^{2x}+3e^x+2}dx$
I made the substitution: $\displaystyle u=e^x$ so $\displaystyle du=e^xdx$
This got me the wrong answer (though my math may have been wrong), is this the correct substitution to make?
$\displaystyle \int\frac{e^{2x}}{e^{2x}+3e^x+2}dx$
I made the substitution: $\displaystyle u=e^x$ so $\displaystyle du=e^xdx$
This got me the wrong answer (though my math may have been wrong), is this the correct substitution to make?
Yes and from there I got:
$\displaystyle \frac{u}{(u+2)(u+1)}=\frac{A}{u+2}+\frac{B}{u+1} \rightarrow u=A(u+1)+B(u+2)$
$\displaystyle u=-2 \text{ then } A=-2$
$\displaystyle u=-1 \text{ then } B=-1$
So...
$\displaystyle \int\frac{-2}{u+2}-\frac{1}{u+1}du$
$\displaystyle -2ln\left|u+2\right|-ln\left|u+1\right|+C$
...and that doesn't seem to be leading to the correct answer, so I think I went wrong somewhere in there.
Another way to find A and B is
u = Au + A + Bu +2B
Now gather the coefficents for each power and make and equation, for example
1 = A + B for power of 1
0 = A + 2B for power of 0
Now you can use substitution or elimination to get your A and B values