Question

Indefinite integral need help pleaseeeeeee

Answer 1

Integrate indefinite integral:
$I=\int\frac{dx}{e^{2x}+3e^x+2}dx$

Solution:
1. use substitution $u=e^x$ 
=>
$du=e^xdx$
=>
$dx=\frac{du}{e^x}$
=>
$dx=\frac{du}{u}$
=>
$I=\int\frac{du}{u(u^2+3u+2)}du$
$=\int\frac{du}{u(u+2)(u+1)}du$
2. decompose into partial fractions
$\frac{1}{u(u+2)(u+1)}$
$=\frac{A}{u}+\frac{B}{u+2}+\frac{C}{u+1}$
where A=1/2, B=1/2, C=-1
$=\frac{1}{2u}+\frac{1}{2(u+2)}-\frac{1}{u+1}$
3. Substitute partial fractions and continue
$I=\int\frac{du}{2u}+\int\frac{du}{2(u+2)}-\int\frac{du}{u+1}$
$=\frac{log(u)}{2}+\frac{log(u+2)}{2}-log(u+1)}$
4. back-substitute u=e^x
$=\frac{log(e^x)}{2}+\frac{log(e^x+2)}{2}-log(e^x+1)}$
$=\frac{x}{2}+\frac{log(e^x+2)}{2}-log(e^x+1)}$

Note: log(x) stands for natural log, and NOT log10(x)