Integrate indefinite integral:
[tex]I=\int\frac{dx}{e^{2x}+3e^x+2}dx[/tex]
Solution:
1. use substitution [tex]u=e^x[/tex]
=>
[tex]du=e^xdx[/tex]
=>
[tex]dx=\frac{du}{e^x}[/tex]
=>
[tex]dx=\frac{du}{u}[/tex]
=>
[tex]I=\int\frac{du}{u(u^2+3u+2)}du[/tex]
[tex]=\int\frac{du}{u(u+2)(u+1)}du[/tex]
2. decompose into partial fractions
[tex]\frac{1}{u(u+2)(u+1)}[/tex]
[tex]=\frac{A}{u}+\frac{B}{u+2}+\frac{C}{u+1}[/tex]
where A=1/2, B=1/2, C=-1
[tex]=\frac{1}{2u}+\frac{1}{2(u+2)}-\frac{1}{u+1}[/tex]
3. Substitute partial fractions and continue
[tex]I=\int\frac{du}{2u}+\int\frac{du}{2(u+2)}-\int\frac{du}{u+1}[/tex]
[tex]=\frac{log(u)}{2}+\frac{log(u+2)}{2}-log(u+1)}[/tex]
4. back-substitute u=e^x
[tex]=\frac{log(e^x)}{2}+\frac{log(e^x+2)}{2}-log(e^x+1)}[/tex]
[tex]=\frac{x}{2}+\frac{log(e^x+2)}{2}-log(e^x+1)}[/tex]
Note: log(x) stands for natural log, and NOT log10(x)
