# understanding sub u

• Dec 2nd 2011, 01:42 PM
delgeezee
understanding sub u
My book has an example that I am having trouble comprehending

Find $\displaystyle \int_{}{} \frac{x}{\sqrt{x+1}} dx$

"Subsititution 1: The composite function $\displaystyle \sqrt{x+1}$ suggests the new variable u = x+1. You might doubt whether this choice will work becuase du=dx and the x in the numerator of the integrand is unnacounted for. But, lets proceed. Letting u = x+1, we have x=u-1, du=dx, and"

$\displaystyle \int_{}{} \frac{x}{\sqrt{x+1}} dx$ = $\displaystyle \int_{}{} \frac{u-1}{\sqrt{u}} du$

....

How should I know to let x= u-1
• Dec 2nd 2011, 01:51 PM
Quacky
Re: understanding sub u
You can integrate $\displaystyle \frac{1}{\sqrt{u}}$, whereas you cannot directly integrate $\displaystyle \frac{1}{\sqrt{x+1}}$

Integration is always a bit of guess and check. One might also consider letting $\displaystyle x=sinh^2(u)$ or letting $\displaystyle x+1=u^2$ because we ultimately want to simplify the radical on the denominator to get it into a form that we can directly integrate. Will my alternative suggestions work? I have no idea (although the former would be particularly ill-advised here!), but they would remove the problem caused by the radical so they would be logical guesses. When dealing with radicals, setting $\displaystyle u$ (or $\displaystyle u^2$) equal to whatever is beneath the radical is usually a wise "first choice". You never "know" they will work for certain, but that's the joy of integration!
• Dec 2nd 2011, 01:52 PM
skeeter
Re: understanding sub u
Quote:

Originally Posted by delgeezee
My book has an example that I am having trouble comprehending

Find $\displaystyle \int_{}{} \frac{x}{\sqrt{x+1}} dx$

"Subsititution 1: The composite function $\displaystyle \sqrt{x+1}$ suggests the new variable u = x+1. You might doubt whether this choice will work becuase du=dx and the x in the numerator of the integrand is unnacounted for. But, lets proceed. Letting u = x+1, we have x=u-1, du=dx, and"

$\displaystyle \int_{}{} \frac{x}{\sqrt{x+1}} dx$ = $\displaystyle \int_{}{} \frac{u-1}{\sqrt{u}} du$

....

How should I know to let x= u-1

not to sound flippant, but experience is the answer to your question. Integration using the method of changing variables comes more naturally after you've done it a few times. Sometimes someone else may see a "clever" substitution that you do not. It all comes with time and practice. Note that u = x+1 allows you to divide out the given quotient so that it only requires a simple power rule antiderivative.
• Dec 2nd 2011, 02:24 PM
delgeezee
Re: understanding sub u
Quote:

Originally Posted by Quacky
You can integrate $\displaystyle \frac{1}{\sqrt{u}}$, whereas you cannot directly integrate $\displaystyle \frac{1}{\sqrt{x+1}}$

Integration is always a bit of guess and check. One might also consider letting $\displaystyle x=sinh^2(u)$ or letting $\displaystyle x+1=u^2$ because we ultimately want to simplify the radical on the denominator to get it into a form that we can directly integrate. Will my alternative suggestions work? I have no idea (although the former would be particularly ill-advised here!), but they would remove the problem caused by the radical so they would be logical guesses. When dealing with radicals, setting $\displaystyle u$ (or $\displaystyle u^2$) equal to whatever is beneath the radical is usually a wise "first choice". You never "know" they will work for certain, but that's the joy of integration!

Yes, letting $\displaystyle x+1=u^2$ is demonstrated in my book as the second method for substituting with this example function. I will heed you advice, "When dealing with radicals, setting $\displaystyle u$ (or $\displaystyle u^2$) equal to whatever is beneath the radical is usually a wise "first choice". "

Quote:

Originally Posted by skeeter
not to sound flippant, but experience is the answer to your question. Integration using the method of changing variables comes more naturally after you've done it a few times. Sometimes someone else may see a "clever" substitution that you do not. It all comes with time and practice. Note that u = x+1 allows you to divide out the given quotient so that it only requires a simple power rule antiderivative.

Agreed, I am will make note of this type of problem and press on.... and when I get more confortable with the substitution method I will return back to it.
• Dec 2nd 2011, 03:22 PM
Re: understanding sub u
Quote:

Originally Posted by delgeezee
My book has an example that I am having trouble comprehending

Find $\displaystyle \int_{}{} \frac{x}{\sqrt{x+1}} dx$

"Substitution 1: The composite function $\displaystyle \sqrt{x+1}$ suggests the new variable u = x+1. You might doubt whether this choice will work because du=dx and the x in the numerator of the integrand is unaccounted for. But, lets proceed. Letting u = x+1, we have x=u-1, du=dx, and"

$\displaystyle \int_{}{} \frac{x}{\sqrt{x+1}} dx$ = $\displaystyle \int_{}{} \frac{u-1}{\sqrt{u}} du$

....

How should I know to let x= u-1

To begin, as Quacky mentioned, it's simpler to have $\displaystyle \sqrt{u}$
under the line.

This leaves you with

$\displaystyle \int{\frac{xdx}{\sqrt{u}}$

and you need to substitute for $\displaystyle xdx$

$\displaystyle x+1=u\Rightarrow\ du=dx$

and also we change from the variable x to u by subtracting 1 from x+1

$\displaystyle x=u-1$

So the rewriting of the entire integral suggests itself after the initial substitution
for the denominator.
• Dec 3rd 2011, 02:33 PM
delgeezee
Re: understanding sub u
Good News, I retried the example problem today and after a while I was able to get the answer. As you all suggested, I found that allowing$\displaystyle u=x+1$ to be the simplest approach. I also now understand why the the numerator changed to$\displaystyle u-1=x$. It was so, to keep the fuction the same, but in terms of u.

I am now attempting, with slight difficulty, to try the other substition example by letting $\displaystyle u= \sqrt{x+1}$

$\displaystyle \int{\frac{x}{\sqrt{x+1}}dx$ = $\displaystyle \int{\frac{x}{u}dx$

Now I change the Numerator in terms of u.

$\displaystyle x= \sqrt{x+1} * \sqrt{x+1} - 1 \Rightarrow u*u-1 \Rightarrow u^2-1$

Now I need to change the variables for the dx in terms of du but I am having troubles.

my first inclination was that since$\displaystyle u = \sqrt{x+1}$, then du = $\displaystyle \frac{1}{2\sqrt{x+1}}$, then replace that x within the du to $\displaystyle \frac{1}{2\sqrt{(u^2-1)+1}}$ = $\displaystyle \frac{1}{2\sqrt{u^2}}$ = $\displaystyle \frac{1}{2u}$

So in the original function, $\displaystyle \int{\frac{x}{\sqrt{x+1}}dx$ , the dx= 1, so I should set the dx = du???
1 = $\displaystyle \frac{1}{2u}$ , multiply both sides to isolute du in terms of u, so finally, du=$\displaystyle 2u$

So Now I have $\displaystyle \int{\frac{u^2-1}{u}*2udu$ which is what the book also has

*********
In regards to finding the du, can I simply say Hey, the derivative of the numerator $\displaystyle u^2-1$ is 2u and i tack it on to the back of the new integrand to get $\displaystyle \int{\frac{u^2-1}{u}*2udu$ ???????
*********

Are these the correct logical steps I should take ?????
• Dec 3rd 2011, 02:45 PM
Quacky
Re: understanding sub u
Nice job so far. I think it's easier to let $\displaystyle u^2=x+1$ rather than to let $\displaystyle u=\sqrt{x+1}$ - both are exactly the same thing, but the first is easier to work with. You can instantly see that $\displaystyle u^2-1=x$ this way. You can also differentiate both sides easily to obtain $\displaystyle 2u~du=dx$ so $\displaystyle dx=2u~du$ instantly, instead of going through a complicated differentiation process.

This leaves you with:

$\displaystyle \int{\frac{u^2-1}{u}}\cdot~2u~du}$

$\displaystyle \int{2u^2-2~du}$
• Dec 3rd 2011, 03:38 PM
Plato
Re: understanding sub u
This a painful thread to follow. (I hate the wide use of u-substitution.)
But in this case it is useful.

If $\displaystyle u=x+1$ then $\displaystyle x=u-1$.

Thus $\displaystyle \frac{x}{\sqrt{x+1}}$ becomes $\displaystyle \frac{u-1}{\sqrt{u}}={u^{1/2}-u^{-1/2}$

That last bit has an easy anti-derivative.
• Dec 3rd 2011, 03:47 PM
Prove It
Re: understanding sub u
Quote:

Originally Posted by Plato
I hate the wide use of u-substitution.

I don't see why - it's a direct method that works in a large number of situations...
• Dec 3rd 2011, 03:49 PM
Quacky
Re: understanding sub u
Quote:

Originally Posted by Plato
This a painful thread to follow. (I hate the wide use of u-substitution.)
But in this case it is useful.

If $\displaystyle u=x+1$ then $\displaystyle x=u-1$.

Thus $\displaystyle \frac{x}{\sqrt{x+1}}$ becomes $\displaystyle \frac{u-1}{\sqrt{u}}={u^{1/2}-u^{-1/2}$

That last bit has an easy anti-derivative.

I agree that u-substitution should be avoided if possible. What alternative would you suggest here?

Quote:

Originally Posted by Prove It
I don't see why - it's a direct method that works in a large number of situations...

With all due respect, it's usually one of the more time-consuming methods; I feel like people rely on it too readily when there may be numerous shorthand approaches available.
• Dec 3rd 2011, 04:01 PM
Prove It
Re: understanding sub u
Quote:

Originally Posted by Quacky
I agree that u-substitution should be avoided if possible. What alternative would you suggest here?

With all due respect, it's usually one of the more time-consuming methods; I feel like people rely on it too readily when there may be numerous shorthand approaches available.

I have to respectfully disagree. I feel it is better to become good at one method instead of having to try to remember numerous "shorthand" methods (in other words, formulas, like \displaystyle \displaystyle \begin{align*} \int{\frac{f'(x)}{f(x)}\,dx} = \ln{\left|f(x)\right|} + C \end{align*} for example). It's also better from a teaching perspective to make sure that students can understand the underlying reasoning behind these "shorthand" methods, which they can only get by becoming good at the u-substitutions.
• Dec 4th 2011, 05:30 AM
Re: understanding sub u
Quote:

Originally Posted by delgeezee
Good News, I retried the example problem today and after a while I was able to get the answer. As you all suggested, I found that allowing$\displaystyle u=x+1$ to be the simplest approach. I also now understand why the the numerator changed to$\displaystyle u-1=x$. It was so, to keep the fuction the same, but in terms of u.

I am now attempting, with slight difficulty, to try the other substition example by letting $\displaystyle u= \sqrt{x+1}$

$\displaystyle \int{\frac{x}{\sqrt{x+1}} \dx$ = $\displaystyle \int{\frac{x}{u}dx$

Now I change the Numerator in terms of u.

$\displaystyle x= \sqrt{x+1} * \sqrt{x+1} - 1 \Rightarrow u*u-1 \Rightarrow u^2-1$

Now I need to change the variables for the dx in terms of du but I am having troubles.

my first inclination was that since$\displaystyle u = \sqrt{x+1}$, then du = $\displaystyle \frac{1}{2\sqrt{x+1}}$, then replace that x within the du to $\displaystyle \frac{1}{2\sqrt{(u^2-1)+1}}$ = $\displaystyle \frac{1}{2\sqrt{u^2}}$ = $\displaystyle \frac{1}{2u}$

So in the original function, $\displaystyle \int{\frac{x}{\sqrt{x+1}}dx$ , the dx= 1, so I should set the dx = du???
1 = $\displaystyle \frac{1}{2u}$ , multiply both sides to isolute du in terms of u, so finally, du=$\displaystyle 2u$

So Now I have $\displaystyle \int{\frac{u^2-1}{u}*2udu$ which is what the book also has

*********
In regards to finding the du, can I simply say Hey, the derivative of the numerator $\displaystyle u^2-1$ is 2u and i tack it on to the back of the new integrand to get $\displaystyle \int{\frac{u^2-1}{u}*2udu$ ???????
*********

Are these the correct logical steps I should take ?????

You are a little bit "heavy-handed" on the calculations.

As mentioned by Prove It and earlier by skeeter,
it is experience with the technique that helps you master it.

$\displaystyle \int{\frac{x}{\sqrt{x+1}}}\ dx$

$\displaystyle u=\sqrt{x+1}$

We require $\displaystyle xdx$ in terms of $\displaystyle u$

$\displaystyle u=\sqrt{x+1}\Rightarrow\ u^2=x+1\Rightarrow\ x=u^2-1$

$\displaystyle \frac{dx}{du}=\frac{d}{du}\left(u^2-1\right)=2u$

$\displaystyle \Rightarrow\ dx=2udu\Rightarrow\ xdx=\left(u^2-1\right)2udu$

$\displaystyle \Rightarrow\int{\frac{x}{\sqrt{x+1}}}\ dx=\int{\frac{\left(u^2-1\right)2u}{u}}\ du=2\int{\left(u^2-1\right)}du$