Integration by subsitution?

Jun 2010
9
1
I am trying to integrate this by parts and not making progress. Is this best done with substitution?

\(\displaystyle \int_1^3 \frac{x}{(x+1)^5}\)
 

Ackbeet

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Jun 2010
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I certainly would. Basically, it allows you to put the variable shift x+1 into the numerator, where it is not raised to the 5th power.
 
Jun 2010
9
1
I certainly would. Basically, it allows you to put the variable shift x+1 into the numerator, where it is not raised to the 5th power.
So you're stating that u substitutes for \(\displaystyle (x+1)^{-5}\) and I integrate from there and subsitute back at the end?
 

Ackbeet

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I would do u = x+1.
 
Dec 2009
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By substitution

\(\displaystyle \int\frac{x}{(x+1)^5}dx = ?\)

Let \(\displaystyle x+1=t\), hence, \(\displaystyle dx=dt\) and \(\displaystyle x=t-1\)

Now, \(\displaystyle \int\frac{x}{(x+1)^5}dx=\int\frac{t-1}{t^5}dt=\int\frac{t}{t^5}dt-\int\frac{1}{t^5}dt=-\frac{t^{-3}}{3}+\frac{t^{-4}}{4}+C\)

And now, back to origins, we get:

\(\displaystyle \int\frac{x}{(x+1)^5}dx=-\frac{(x+1)^{-3}}{3}+\frac{(x+1)^{-4}}{4}+C\)
 

Ackbeet

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Jun 2010
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Your original integral had limits. If you have limits, there's no need to get back to the original variable of integration, so long as you transform the limits into the new variable. Also, I think your integrations are incorrect. Don't skip so many steps!
 

Ackbeet

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Jun 2010
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Oh, I must have been seeing different numbers in the denominator. It's been a long day!

Still need to plug in limits, though. Perhaps we could let cwashburn do that?
 
Jun 2010
9
1
Per your suggestion Ackbeet I substituted y for x+1.

That led me to x = y-1 and so I had \(\displaystyle (y-1)*y^{-5}\). Multiplying through I get \(\displaystyle \int y^{-4} - \int y^{-5}\).

So after integrating I have \(\displaystyle \int -\frac{1}{3y^{3}} - \int-\frac{1}{4y^{4}}\)

I've been balancing with wolframalpha against the result I set up as a formula in Excel.

Interestingly my integration \(\displaystyle \frac{-1}{3y^{3}}\mid_1^3 - \frac{-1}{4y^{4}}\mid_1^3\) works, but on the interval of 2 to 4, versus 1 to 3, which are the original boundaries.

Do I need to increase the intervals by 1 because prior to substitution I had x+1? Is that why 2 to 4 works out to the right answer on \(\displaystyle \frac{-1}{3y^{3}}\mid_{y+1}^{x+1} - \frac{-1}{4y^{4}}\mid_{y+1}^{x+1}\) where x and y are my original limits?