How do I evaluate $intdx/(e^(x)(3e^(x)+2))$ ?

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Answer 1 · 2018-06-18T15:55:49

The answer is $=-1/(2e^x)-3/4x+3/4ln(3e^x+2)+C$

Let $e^x=u$ , $=>$ , $du=e^xdx$

Therefore, the integral is

$I=int(dx)/(e^x(3e^x+2))=int(du)/(e^(2x)(3e^x+2))$

$=int(du)/(u^2(3u+2))$

Perform the decomposition into partial fractions

$1/(u^2(3u+2))=A/u^2+B/u+C/(3u+2)$

$=(A(3u+2)+Bu(3u+2)+Cu^2)/(u^2(3u+2))$

The denominator is the same, compare the numerators

$1=A(3u+2)+Bu(3u+2)+Cu^2$

Let $u=0$ , $=>$ , $1=2A$ , $=>$ , $A=1/2$

Coefficients of $u$

$0=3A+2B$ , $=>$ , $B=-3/2A=-3/4$

Coefficients of $u^2$

$0=3B+C$ , $=>$ , $C=-3B=9/4$

Therefore,

$1/(u^2(3u+2))=(1/2)/u^2+(-3/4)/u+(9/4)/(3u+2)$

So,

$int(du)/(u^2(3u+2))=1/2int(du)/u^2-3/4int(du)/u+9/4int(du)/(3u+2)$

$=-1/(2u)-3/4lnu+3/4ln(3u+2)$

$=-1/(2e^x)-3/4ln(e^x)+3/4ln(3e^x+2)+C$

$=-1/(2e^x)-3/4x+3/4ln(3e^x+2)+C$

How do I evaluate intdx/(e^(x)(3e^(x)+2))intdx/(e^(x)(3e^(x)+2))?