Evaluate the integral with partial fraction decomposition

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MarkSA

Junior Member
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Sep 8, 2007
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243
Hello,

1) Evaluate the integral (using partial fraction decomposition first):

integral of: x^3/([x + 1]^3) dx

I decomposed it... to:
A/(x + 1) + B/(x + 1)^2 + C/(x + 1)^3
x^3 = x^2(A) + x(2A + B) + (A + B + C) gets me A = 0, B = 0, C = 0? But that can't be right, because that makes the entire integral 0.

Any suggestions?
 
Because I have such disdain for the sort of problem here is the answer.
Mind you, this is done with a CAS that cost less that $40 for Windrows users.
 

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Do you know how the partial fraction is arrived at?

I don't understand where the "1 -" in front comes from. The rest of the denominators look similar to the way I attempted to do it.
 
MarkSA said:
Do you know how the partial fraction is arrived at?
Click to expand...
To learn how to do partial-fraction decomposition, try studying some of the many great lessons available online! :wink:

MarkSA said:
I don't understand where the "1 -" in front comes from.
Click to expand...
To do partial-fraction decomposition, (you will learn that) you need to start with a "proper" polynomial fraction. Since the original fraction, in your case, was "improper" (that is, the degree of the denominator was not strictly larger than the degree of the numerator), one must first apply long division to find the non-fractional portion of the expression.

In effect, you have to turn the rational expression into a "mixed number" sort of thing, similar to converting 5/3 to 1[sup:1lhrum2g]2[/sup:1lhrum2g]/[sub:1lhrum2g]3[/sub:1lhrum2g].

Eliz.
 
Woops! That's the second time I forgot about having to divide it out today... no wonder I had so much trouble with this one. I will try it again now, thanks
 
MarkSA said:
Hello,

1) Evaluate the integral (using partial fraction decomposition first):

integral of: x^3/([x + 1]^3) dx

I decomposed it... to:
A/(x + 1) + B/(x + 1)^2 + C/(x + 1)^3
x^3 = x^2(A) + x(2A + B) + (A + B + C) gets me A = 0, B = 0, C = 0? But that can't be right, because that makes the entire integral 0.

Any suggestions?
Click to expand...

This one is "particularly"simple:

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

\(\displaystyle = \,\,(\frac{x}{(x+1)})^3\)

\(\displaystyle = \,\,(1 - \frac{1}{(x+1)})^3\)

Now simply expand the cubed function using

\(\displaystyle (a-b)^3\, = \, a^3 \, - \, 3a^2b\, + \, 3ab^2\, -\, 1\)
 
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