Can someone help me with this integration by parts?

- AmTran_Bus

Can someone help me with this integration by parts?

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- AmTran_Bus

|dw:1433873040803:dw|

- AmTran_Bus

|dw:1433873069974:dw|

- AmTran_Bus

That is supposed to say du= 2r dr

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## More answers

- dumbcow

you forgot the sqrt around the "4+r^2"
actually i dont think you need int by parts here .... just the substitution
--> u = 4+r^2
du = 2r dr

- AmTran_Bus

Hum. This problem is from the integration by parts chapter in the Stewart book :(

- AmTran_Bus

But ok. Lets do it your way! Can you guide me through it?

- dumbcow

sure ... double check you copied the question correctly though

- AmTran_Bus

Ok. I did.

- dumbcow

after substitution you get
\[\frac{(u-4) r}{\sqrt{u}} * \frac{du}{2r}\]

- dumbcow

the "r" cancels leaving
\[\frac{1}{2} \int\limits \frac{u-4}{\sqrt{u}} du\]

- AmTran_Bus

Ok, can you explain the after substitution step? Is u=4+r^2?

- dumbcow

yes so since u = 4 +r^2 , if you solve for r^2 you get r^2 = u-4
--> r^3 = r^2 *r = (u-4)r

- AmTran_Bus

Got ya. Thanks.

- AmTran_Bus

Don't we have to solve for out limits?

- AmTran_Bus

*our

- dumbcow

yes but first you must find integral .... then plug in limits at the end

- AmTran_Bus

##### 1 Attachment

- dumbcow

can you finish the integral?
\[\frac{u-4}{\sqrt{u}} = u^{1/2} -4u^{-1/2}\]

- phi

FYI if you want to use integration by parts, notice that to integrate
\[ (4+r^2)^{-\frac{1}{2} } \]
Taking that as a hint, write the problem as
\[ \int \frac{r^3}{\sqrt{4+r^2}} dr = \int r^2 (4+r^2)\ r\ dr \]
and let
\[ u= r^2 \\ dv = (4+r^2)\ r\ dr \]

- AmTran_Bus

Ok. So the (4+r^2)^(-1/2) was basically put on top now, right?

- phi

yes, only because I don't like square roots.

- phi

to integrate
\[ (4+r^2)^{-\frac{1}{2} } \]
we need 2 r dr

- AmTran_Bus

So we havae|dw:1433877516461:dw|

- AmTran_Bus

Im just trying to get to what you did to be sure I get it.

- phi

so split the r^3 into r^2 and r (and put the r next to the dr)

- AmTran_Bus

Ok. r^2 *r

- phi

yes. so it should look like this
\[ \int \frac{r^3}{\sqrt{4+r^2}} dr = \int r^2 (4+r^2)^{-\frac{1}{2}}\ r\ dr\]

- AmTran_Bus

Very nice. So now I can just do the regular U and dv stuff.

- phi

yes, u= r^2
dv = what is left over

- AmTran_Bus

Thank you so very much. I still have just a small issue with how you got|dw:1433877820455:dw|

- AmTran_Bus

Sorry for that mess. And for asking you so much.

- AmTran_Bus

I get how you brought the sqrt up top.

- AmTran_Bus

|dw:1433877903541:dw|

- phi

you don't have to change the square root to an exponent, but I find it easier to think about|dw:1433877934816:dw|

- phi

\[ \frac{1}{x}= x^{-1} \]
\[ \sqrt{x} = x^\frac{1}{2} \]
\[ \frac{1}{\sqrt{x}}=\frac{1}{x^\frac{1}{2}}=x^{-\frac{1}{2}} \]

- AmTran_Bus

Thanks so much phi. I am really glad you showed me how to do it this way.

- phi

You still have to do the integral

- AmTran_Bus

I know, from all the U= and DV= stuff. But I'm pretty good with that.

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