Integration by substitution method involving e

[James10492]

Hi all, I am a bit stuck on this problem and would appreciate some pointers. Here is the problem and how I have attempted it so far.

"Show that

$$\int_{\ln4}^{\ln3} \frac{e^{4x}}{e^x - 2} = \frac{a}{b} +c \ln d$$
where a, b, c and d are integers to be found. Use

$$u^2 = e^x - 2$$ "


To evaluate the integral and prove the statement, find the integral, which I have tried to do like this:


$$2udu = e^x dx$$
and

$$e^x = u^2 + 2$$
so

$$dx = \frac{2udu}{u^2+2}$$
now

$$\int \frac{e^{4x}}{e^x - 2}dx = \frac{e^{4x}}{u^2} \frac{2udu}{u^2+2}$$

$$e^x = u^2 +2$$
$$e^{4x} = (e^x)^4 = (u^2 + 2)^4$$
so

$$\int \frac{2u(u^2 +2)^4}{u^2(u^2+2)}du$$
cancelling,

$$\int \frac {2(u^2 +2)^3}{u}du$$
So the substitution has not really led to a simplified expression and I do not know how to continue from here. Is there something better I could have done?

 

[Cubist]

So the substitution has not really led to a simplified expression ...

It looks messy, but think about it, once expanded it'll be easy to integrate! It will basically be a polynomial in "u" with a single "1/u" term added on the end. Moreover, can you see that all the powers of u will end up odd after the expansion? Will they still be odd after integration? Will this fit with the new limits? Therefore I'd keep going (what you have done is correct so far)

EDIT: not literally "1/u" but it will be some constant c/u

 

[Deleted member 4993]

Hi all, I am a bit stuck on this problem and would appreciate some pointers. Here is the problem and how I have attempted it so far.

"Show that

$$\int_{\ln4}^{\ln3} \frac{e^{4x}}{e^x - 2} = \frac{a}{b} +c \ln d$$
where a, b, c and d are integers to be found. Use

$$u^2 = e^x - 2$$ "


To evaluate the integral and prove the statement, find the integral, which I have tried to do like this:


$$2udu = e^x dx$$
and

$$e^x = u^2 + 2$$
so

$$dx = \frac{2udu}{u^2+2}$$
now

$$\int \frac{e^{4x}}{e^x - 2}dx = \frac{e^{4x}}{u^2} \frac{2udu}{u^2+2}$$

$$e^x = u^2 +2$$
$$e^{4x} = (e^x)^4 = (u^2 + 2)^4$$
so

$$\int \frac{2u(u^2 +2)^4}{u^2(u^2+2)}du$$
cancelling,

$$\int \frac {2(u^2 +2)^3}{u}du$$
So the substitution has not really led to a simplified expression and I do not know how to continue from here. Is there something better I could have done?

Expand numerator using (a + b)^3 = a^3 + b^3 + 3ab(a+b)

 

[Steven G]

Although you should know how to cube a sum you can do the following.

$\displaystyle \dfrac {(u^2+2)^3}{u} = \dfrac {u^2+2}{u}(u^2+2)^2 = (u+\dfrac{2}{u})(u^4+4u^2+4) = ...$

 

[Steven G]

Hi all, I am a bit stuck on this problem and would appreciate some pointers. Here is the problem and how I have attempted it so far.

"Show that

$$\int_{\ln4}^{\ln3} \frac{e^{4x}}{e^x - 2} = \frac{a}{b} +c \ln d$$
where a, b, c and d are integers to be found. Use

$$u^2 = e^x - 2$$ "


To evaluate the integral and prove the statement, find the integral, which I have tried to do like this:


$$2udu = e^x dx$$
and

$$e^x = u^2 + 2$$
so

$$dx = \frac{2udu}{u^2+2}$$
now

$$\int \frac{e^{4x}}{e^x - 2}dx = \frac{e^{4x}}{u^2} \frac{2udu}{u^2+2}$$

$$e^x = u^2 +2$$
$$e^{4x} = (e^x)^4 = (u^2 + 2)^4$$
so

$$\int \frac{2u(u^2 +2)^4}{u^2(u^2+2)}du$$
cancelling,

$$\int \frac {2(u^2 +2)^3}{u}du$$
So the substitution has not really led to a simplified expression and I do not know how to continue from here. Is there something better I could have done?

Maybe not in this case, but it is possible that making a 2nd substitution will help. That is always a possibility that you should consider.