Integration of (e^x)dx/ (e^(2x) +5e^(x) + 6)

[k0hana27]

I was really confuse with this problem because after i simplify the denominator which is (e^x+2)(e^x+3) and use partial fraction i get A=3 and B=-2. Whenever I get the derivative of my answer it doens't match with the given (e^x)dx/ (e^(2x) +5e^(x) + 6). PLEASE help me

 

[xiaoB]

Try using this way: Let y=e^x

 

[arildno]

I was really confuse with this problem because after i simplify the denominator which is (e^x+2)(e^x+3) and use partial fraction i get A=3 and B=-2. Whenever I get the derivative of my answer it doens't match with the given (e^x)dx/ (e^(2x) +5e^(x) + 6). PLEASE help me

Your partial fraction decomposition is certainly correct; so your problem lies in the following integration.

I suggest you use, from the outset, the substitution pointed out to you, and take the fractional decomposition afterwards.

 

[ELog]

You're rushing into partial fractions too quickly. First substitute $$u=e^x$$ so that $$du=e^xdx$$. Partial fractions will then yield the correct integration.

 

[HallsofIvy]

It doesn't matter whether you substitute for $e^x$ before or after the "partial fractions"- you would get the same answer. The real problem is that K0hana27 hasn't shown us how he integrated or what answer he got.

 

[arildno]

It doesn't matter whether you substitute for $e^x$ before or after the "partial fractions"- you would get the same answer. The real problem is that K0hana27 hasn't shown us how he integrated or what answer he got.

Sure, it doesn't matter.

Except for the loathsome fact that if you afterwards performs the substitution, you'll most likely to do yet another partial fractions decomposition.
On both the two fractional terms previously gained.

It is easier to make the obvious substitution first.