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#