# Thread: finding the exact value of an integral which contains log and e

1. ## finding the exact value of an integral which contains log and e

find the exact value of:
ln7

⌡ int e^(6x)(e^(6x) + 10)^7/2 dx
ln5

I have a brief idea of how to solve this, but I'm not too sure about the specific parts.
You can make e^6x=u, differentiate this to find dx, but I'm not too sure about the new limits and how to solve the rest. Is there any easier way to solve this, other than the substitution method?

any help would be great, thanks

2. Perhaps with just a little more experience, you will see that this one is screaming for a substitution. Try u = e^(6x) + 10.

3. Hi!

I remember at school there used to be this formula that we blindly used as a tool whenever something like this popped up, then we learnt that it's in fact a form substitution method:

$\int u'(x) . u^n (x) dx = \int u^n du$.

Can you know why? Just remember that $u' =\frac{du}{dx}$

And remember, by this so called change of variable method we're now integrating w.r.t. u and not x. So if $u = 2x+1$ and the limits are x = 0, 1, the limits of u are 1, 3.

That plus TKHunny's hint, can you complete this?

4. Originally Posted by rebghb
Hi!

I remember at school there used to be this formula that we blindly used as a tool whenever something like this popped up, then we learnt that it's in fact a form substitution method:
$\int u'(x) . u^n (x) dx = \int u^n du$.

⌡ u'(x) . u^n(x) dx =

⌡ u^n du
Can you know why? Just remeber that $u' = \frac{du}{dx}$ u' = du/dx.
And remeber, by this so called change of variable method we're now integrating w.r.t. $u$ and not $x$. So if $u = 2x+1$ and the limits are $x=0, 1$ x = 0, 1, the limits of $u$ are $1, 3$ 1, 3.
That plus TKHunny's hint, can you complete this?

Editing Note: Sorry, there's a problem with the LaTeX compiler
$\int_{ln(5)}^{ln(7)} e^{6x}(e^{6x} + 10)^{7/2}~dx$

Let u = e^{6x} + 10. Your lower limit for x is ln(5), so plug x = ln(5) into the u formula to get the lower limit for u...

For the rest of it, you can easily solve for e^{6x} from the u equation. So you can find the integrand in terms of u. Finally you can find du = ( ) dx. Solve for dx and plug that into the integral. You will get

$\int_{5^6 + 10}^{7^6 + 10}(u - 10)u^{7/2}~\frac{1}{6} \cdot \frac{du}{u - 10}$

Can you take it from there?

-Dan

5. Originally Posted by topsquark
$\int_{ln(5)}^{ln(7)} e^{6x}(e^{6x} + 1)^{7/2}~dx$
You will get

$\int_{5^6 + 10}^{7^6 + 10}(u - 10)u^{7/2}~\frac{1}{6} \cdot \frac{du}{u - 10}$
I get everything else, but would you please explain how you found that equation in terms of u? thanks

6. Let u = e^{6x} + 10. Then, in the integrand

$e^{6x}(e^{6x} + 10)^{7/2} = (u - 10)u^{7/2}$

Now we need du in terms of dx.
$u = e^{6x} + 10 \implies du = 6e^{6x}~dx$

We need to put the e^{6x} in terms of u. But we know that e^{6x} = u - 10. So
$u = e^{6x} + 10 \implies du = 6e^{6x}~dx = 6(u - 10)dx$

So
$dx = \frac{1}{6}\frac{du}{u - 10}$

Now sub dx into the original integral and the result follows.

-Dan

7. Hello all!
Oh dear, topsquark, I don't know why you want to go there! Since $e^{6x} + 10$ isn't just any regular function, why not going as follows:
$u = e^{6x}+10$ and noticing that $6e^{6x}=\frac{du}{dx}$. Then:
$\displaysyle \int e^{6x} (e^{6x}+10)^{7/2}dx= \int \frac{du}{dx}u^{7/2}dx$
= $\displaystyle \int u^{7/2} du = (2/7) u^{9/2}$ between your 2 modified limits.

Other version:

⌡(1/6). u'(x) . u^n(x) dx =

⌡(1/6) u^(7/2) du = (1/6) / (9/2) u ^ (9/2) = (1/27) . (e^(6x) + 10) ^ (9/2)

Somebody please do something with the LaTeX compiler!

8. Originally Posted by rebghb
Hello all!
Oh dear, topsquark, I don't know why you want to go there! Since $e^{6x} + 10$ isn't just any regular function, why not going as follows:
$u = e^{6x}+10$ and noticing that $6e^{6x}=\frac{du}{dx}$. Then:
$\displaysyle \int e^{6x} (e^{6x}+10)^{7/2}dx= \int \frac{du}{dx}u^{7/2}dx$
= $\displaystyle \int u^{7/2} du = (2/7) u^{9/2}$ between your 2 modified limits.
(shrugs) It's not like it's that hard. And it's a simple and direct application of the substitution method. But you like your way and I like mine. No problems.

Originally Posted by rebghb
Somebody please do something with the LaTeX compiler!
See post 8 in this thread.

-Dan