# Thread: Integration formula problem

1. ## Integration formula problem

Hi all,

I'm having difficulty manipulating a problem so it fits the equation required to use the formula.

I've been asked by using equation

$\displaystyle \int(f(x))^{n}f'(x)dx=\frac{1}{n+1}(f(x))^{n+1}+c$
show that

$\displaystyle \int\frac{e^{x}+e^{-2x}}{(2e^{x}-e^{-2x})^{\frac{3}{2}}}dx=-(2e^{x}-e^{-2x})^{-\frac{1}{2}}+c$

please don't reveal too much to me, but how can I manipulate

$\displaystyle \int\frac{e^{x}+e^{-2x}}{(2e^{x}-e^{-2x})^{\frac{3}{2}}}dx$

so it is of the form

$\displaystyle \int(f(x))^{n}f'(x)dx$

many thanks.

2. ## Re: Integration formula problem

Originally Posted by Hydralisk
Hi all,

I'm having difficulty manipulating a problem so it fits the equation required to use the formula.

I've been asked by using equation

$\displaystyle \int(f(x))^{n}f'(x)dx=\frac{1}{n+1}(f(x))^{n+1}+c$
show that

$\displaystyle \int\frac{e^{x}+e^{-2x}}{(2e^{x}-e^{-2x})^{\frac{3}{2}}}dx=-(2e^{x}-e^{-2x})^{-\frac{1}{2}}+c$

please don't reveal too much to me, but how can I manipulate

$\displaystyle \int\frac{e^{x}+e^{-2x}}{(2e^{x}-e^{-2x})^{\frac{3}{2}}}dx$

so it is of the form

$\displaystyle \int(f(x))^{n}f'(x)dx$

many thanks.
Firstly, what is the derivative of

$\displaystyle 2e^x-e^{-2x}\;\;\;?$

3. ## Re: Integration formula problem

Consider $\displaystyle \displaystyle \frac{1}{f^{\frac{3}{2}}}= f^{\frac{-3}{2}}$ and $\displaystyle \displaystyle\frac{-3}{2}+1 = \frac{-1}{2}$

4. ## Re: Integration formula problem

I've got

$\displaystyle \frac{d}{dx}2e^{x}-e^{2x}=2e^{x}-2e^{2x}$

but unfortunately I can't see what to do with it now?

5. ## Re: Integration formula problem

Originally Posted by pickslides
Consider $\displaystyle \displaystyle \frac{1}{f^{\frac{3}{2}}}= f^{\frac{-3}{2}}$ and $\displaystyle \displaystyle\frac{-3}{2}+1 = \frac{-1}{2}$
I'm sorry but this has gone "over my head", I think I may need a bit more "help an idiot" style tips.

I feel a bit stupid asking for help, but I really can't seem to even make a start on it.

6. ## Re: Integration formula problem

Your derivative is incorrect. $\displaystyle \displaystyle \frac{d}{dx}(2e^x - e^{-2x}) = 2e^x + 2e^{-2x} = 2(e^x + e^{-2x})$.

Rewrite your integral as

$\displaystyle \displaystyle \frac{1}{2}\int{2(e^{x} + e^{-2x})(2e^x - e^{-2x})^{-\frac{3}{2}}\,dx}$

7. ## Re: Integration formula problem

Originally Posted by pickslides
Consider $\displaystyle \displaystyle \frac{1}{f^{\frac{3}{2}}}= f^{\frac{-3}{2}}$ and $\displaystyle \displaystyle\frac{-3}{2}+1 = \frac{-1}{2}$
Does

$\displaystyle \int\frac{e^{x}+e^{-2x}}{(2e^{x}-e^{-2x})^{\frac{3}{2}}}dx=\int(e^{x}+e^{-2x})(2e^{x}-e^{-2x})^{-3/2}dx$

Am I on the right track? If I can get rid of the fraction I may be able to figure it out from there.

8. ## Re: Integration formula problem

Originally Posted by Prove It
Your derivative is incorrect. $\displaystyle \displaystyle \frac{d}{dx}(2e^x - e^{-2x}) = 2e^x + 2e^{-2x} = 2(e^x + e^{-2x})$.

Rewrite your integral as

$\displaystyle \displaystyle \frac{1}{2}\int{2(e^{x} + e^{-2x})(2e^x - e^{-2x})^{-\frac{3}{2}}\,dx}$
I "think" I may be OK from here on.

Many many thanks,

I've got to leave for work now, but tonight I'll give it a good try to carry on through it.

Thanks again to all of you.

9. ## Re: Integration formula problem

Originally Posted by Hydralisk
I've got

$\displaystyle \frac{d}{dx}\left(2e^{x}-e^{2x}\right)=2e^{x}-2e^{2x}$

but unfortunately I can't see what to do with it now?
Yes, that derivative is correct,
but you should have had a negative exponent on the 2nd term initially.
Then you will see that this derivative is

$\displaystyle 2\left(e^x+e^{-2x}\right)$

Then use

$\displaystyle \frac{1}{\left(2e^x-e^{-2x}\right)^{\frac{3}{2}}}=\frac{\left(2e^x-e^{-2x}\right)^0}{\left(2e^x-e^{-2x}\right)^{\frac{3}{2}}}$

$\displaystyle =\left(2e^x-e^{-2x}\right)^{-\frac{3}{2}}$

10. ## Re: Integration formula problem

Thank you all so much,

I eventually got through it using everything you said above.