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

Question

Ezekiel Alexander · Accepted Answer

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

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

Consequently, the integral is

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

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

Divide the breakdown 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))#

Compare the numerators; the denominator is the same.

#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#

Consequently,

#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#

Luke Alvarez · Answer

undefined To evaluate the integral ∫ dx/(e^x(3e^x + 2)), you can use substitution. Let u = e^x. Then, du = e^x dx. Rewrite the integral in terms of u:

∫ du / (u * (3u + 2))

Now, use partial fraction decomposition to split the fraction:

1/(u * (3u + 2)) = A/u + B/(3u + 2)

Find A and B by equating numerators:

1 = A(3u + 2) + Bu

Solve for A and B. Then integrate each term separately. After integrating, substitute back u = e^x to get the final result.