understanding sub u

**delgeezee** · December 2nd 2011, 01:42 PM

My book has an example that I am having trouble comprehending

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

"Subsititution 1: The composite function $\sqrt{x+1}$ 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"

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

....

How should I know to let x= u-1

**Quacky** · December 2nd 2011, 01:51 PM

You can integrate $\frac{1}{\sqrt{u}}$ , whereas you cannot directly integrate $\frac{1}{\sqrt{x+1}}$

Integration is always a bit of guess and check. One might also consider letting $x=sinh^2(u)$ or letting $x+1=u^2$ 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 $u$ (or $u^2$ ) 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!

**skeeter** · December 2nd 2011, 01:52 PM

Originally Posted by delgeezee

My book has an example that I am having trouble comprehending

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

"Subsititution 1: The composite function $\sqrt{x+1}$ 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"

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

....

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.

**delgeezee** · December 2nd 2011, 02:24 PM

Originally Posted by Quacky

You can integrate $\frac{1}{\sqrt{u}}$ , whereas you cannot directly integrate $\frac{1}{\sqrt{x+1}}$

Integration is always a bit of guess and check. One might also consider letting $x=sinh^2(u)$ or letting $x+1=u^2$ 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 $u$ (or $u^2$ ) 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 $x+1=u^2$ 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 $u$ (or $u^2$ ) equal to whatever is beneath the radical is usually a wise "first choice". "

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.

**Archie Meade** · December 2nd 2011, 03:22 PM

Originally Posted by delgeezee

My book has an example that I am having trouble comprehending

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

"Substitution 1: The composite function $\sqrt{x+1}$ 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"

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

....

How should I know to let x= u-1

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

This leaves you with

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

and you need to substitute for $xdx$

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

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

$x=u-1$

So the rewriting of the entire integral suggests itself after the initial substitution
for the denominator.

**delgeezee** · December 3rd 2011, 02:33 PM

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 $u=x+1$ to be the simplest approach. I also now understand why the the numerator changed to $u-1=x$ . 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 $u= \sqrt{x+1}$

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

Now I change the Numerator in terms of u.

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

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

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

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

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

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

**Quacky** · December 3rd 2011, 02:45 PM

Nice job so far. I think it's easier to let $u^2=x+1$ rather than to let $u=\sqrt{x+1}$ - both are exactly the same thing, but the first is easier to work with. You can instantly see that $u^2-1=x$ this way. You can also differentiate both sides easily to obtain $2u~du=dx$ so $dx=2u~du$ instantly, instead of going through a complicated differentiation process.

This leaves you with:

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

Leading to:

$\int{2u^2-2~du}$

**Plato** · December 3rd 2011, 03:38 PM

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

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

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

That last bit has an easy anti-derivative.

**Prove It** · December 3rd 2011, 03:47 PM

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...

**Quacky** · December 3rd 2011, 03:49 PM

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 $u=x+1$ then $x=u-1$ .

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

That last bit has an easy anti-derivative.

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

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.

**Prove It** · December 3rd 2011, 04:01 PM

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 $\displaystyle \begin{align*} \int{\frac{f'(x)}{f(x)}\,dx} = \ln{\left|f(x)\right|} + C \end{align*}$ 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.

**Archie Meade** · December 4th 2011, 05:30 AM

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 $u=x+1$ to be the simplest approach. I also now understand why the the numerator changed to $u-1=x$ . 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 $u= \sqrt{x+1}$

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

Now I change the Numerator in terms of u.

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

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

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

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

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

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.

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

$u=\sqrt{x+1}$

We require $xdx$ in terms of $u$

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

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

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

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

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