1. ## Integral Help

int(3*(x^2+9)^10)

How can you do this using U sub?

2. Originally Posted by anywho
int(3*(x^2+9)^10)

How can you do this using U sub?
$\displaystyle \int 3(x^2+9)^{10}dx$

This is not a candidate for regular subsitution. There is no x in front of our bracket. In order to use regular substution whatever is in front of the bracket must be 1 order less then the term(s) in the bracket. In this case we would need an X.

You cannt evaluate via this method.

What you can do is integrate by Parts and then on your

$\displaystyle \int VdU$

Integral let

$\displaystyle p=x^2 + 9 --> x^2 = p -9$

This should work to get this to a nice integral for yeah

3. Originally Posted by AllanCuz
$\displaystyle \int 3(x^2+9)^{10}dx$

This is not a candidate for regular subsitution. There is no x in front of our bracket. In order to use regular substution whatever is in front of the bracket must be 1 order less then the term(s) in the bracket. In this case we would need an X.

You cannt evaluate via this method.

What you can do is integrate by Parts and then on your

$\displaystyle \int VdU$

Integral let

$\displaystyle p=x^2 + 9 --> x^2 = p -9$

This should work to get this to a nice integral for yeah
Are you sure..?

Obviously we must set u (as in u from uv - int vu') as $\displaystyle (x^2 + 9)^{10}$ since if we knew how to integrate that we wouldn't need to use integration by parts...

So we get...

$\displaystyle u = (x^2 + 9)^{10}$

$\displaystyle u' = 20x(x^2 + 9)^9$

$\displaystyle v = 3x$

Hence we get...

$\displaystyle 3x(x^2 + 9)^{10} - \int 60x^2(x^2 + 9)^9 dx$ ...

Setting $\displaystyle x^2 = p-9$ gives...

$\displaystyle x = \sqrt{p-9}, dx = \frac{1}{2} (p-9)^{-1/2} dp$

Hence ignoring the uv part the integral part becomes...

$\displaystyle \int \frac{60(p-9)p^9}{2 \sqrt{p-9}} dp$

Is this easy? I can't see an easy way..?

I'm actually stumped by this question as well...

Perhaps we need a double substitution or even a trig sub in the above integral...

Are you sure..?

Obviously we must set u (as in u from uv - int vu') as $\displaystyle (x^2 + 9)^{10}$ since if we knew how to integrate that we wouldn't need to use partial integration...

So we get...

$\displaystyle u = (x^2 + 9)^{10}$

$\displaystyle u' = 20x(x^2 + 9)^9$

$\displaystyle v = 3x$

Hence we get...

$\displaystyle 3x(x^2 + 9)^{10} - \int 60x^2(x^2 + 9)^9 dx$ ...

Setting $\displaystyle x^2 = p-9$ gives...

$\displaystyle x = \sqrt{p-9}, dx = \frac{1}{2} (p-9)^{-1/2} dp$

Hence ignoring the uv part the integral part becomes...

$\displaystyle \int \frac{60(p-9)p^9}{2 \sqrt{p-9}} dp$

Is this easy? I can't see an easy way..?

I'm actually stumped by this question as well...

Perhaps we need a double substitution or even a trig sub in the above integral...
Yeah I didn't fully write it out on paper so I thought it might simpify LOL
You could do trig in your integral or we could find a reduction of order, seeing as this will continue to reduce (in terms of the variables). But yeah, i don't see this as being overly easy at all...

Edit--What you did was essentially the method of Integration by Parts. When someone says U subsitution I usually take it to mean U=Inside the bracket, which in this case wasn't possible.

Are you sure..?

Obviously we must set u (as in u from uv - int vu') as $\displaystyle (x^2 + 9)^{10}$ since if we knew how to integrate that we wouldn't need to use integration by parts...

So we get...

$\displaystyle u = (x^2 + 9)^{10}$

$\displaystyle u' = 20x(x^2 + 9)^9$

$\displaystyle v = 3x$

Hence we get...

$\displaystyle 3x(x^2 + 9)^{10} - \int 60x^2(x^2 + 9)^9 dx$ ...

Setting $\displaystyle x^2 = p-9$ gives...

$\displaystyle x = \sqrt{p-9}, dx = \frac{1}{2} (p-9)^{-1/2} dp$

Hence ignoring the uv part the integral part becomes...

$\displaystyle \int \frac{60(p-9)p^9}{2 \sqrt{p-9}} dp$

Is this easy? I can't see an easy way..?

I'm actually stumped by this question as well...

Perhaps we need a double substitution or even a trig sub in the above integral...
Yes. This question is not as easy and as small as it looks. Here's Wolfram Alpha for you:

integrate&#x28;3&#x2a;&#x28;x&#x5e;2&#x2b;9&#x29;& #x5e;10&#x29; - Wolfram|Alpha

6. Originally Posted by harish21
Yes. This question is not as easy and as small as it looks. Here's Wolfram Alpha for you:

integrate&#x28;3&#x2a;&#x28;x&#x5e;2&#x2b;9&#x29;& #x5e;10&#x29; - Wolfram|Alpha
I think it's easier than that. That is just the binomial expansion being integrated I think. I.e. (I think this is the right expansion...)
$\displaystyle 3 \int (x^2 + 9)^{10} dx = 3 \int \sum_{j=0}^{10} x^{2j} 9^{10-j} {10 \choose j} dx$

So we could just integrate that to get...

$\displaystyle 3\sum_{j=0}^{10} \frac{x^{2j+1} 9^{10-j}}{2j+1} {10 \choose j}$

But I believe there will be a more elegant solution...

$\displaystyle 3 \int (x^2 + 9)^{10} dx = 3 \int \sum_{j=0}^{10} x^{2j} 9^{10-j} {10 \choose j} dx$
$\displaystyle 3\sum_{j=0}^{10} \frac{x^{2j+1} 9^{10-j}}{2j+1} {10 \choose j}$