I put this integration problem on wolfram alpha
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# I put this integration problem on wolfram alpha

[From: ] [author: ] [Date: 11-05-25] [Hit: ]
Done!When x = 2,When x = -2,Finally knowing that all of the x^2 terms must add to 0,......
but can't figure out HOW they derived the partial fractions (the 3rd step)?????
http://www.wolframalpha.com/input/?i=int…
incase the link doesn't work, my original problem was: integral of 2/ (x+2)^2(2-x) dx

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2 /(x + 2)^2(2 - x) = A/(x+2) + B/(x+2)^2 + C/(2 - x)

2 = A(x + 2)(2 - x) + B(2 - x) + C(x+2)^2

2 = A(4 - x^2) + 2B - Bx + C(x^2 + 4x + 4)

2 = x^2(-A + C) + x( - B + 4C) + 4A + 2B + 4C

- A + C = 0 ==> A = C
-B + 4C = 0 ==> B = 4C
4A + 2B + 4C = 2 ==> 4C + 8C + 4C = 2 ==> C = 1/8

A = 1/8, B = 1/2 and C = 1/8

2 /(x + 2)^2(2 - x) = (1/8) { 1/(x+2) } + (1/2) { 1/(x+2)^2 } + (1/8) { 1/(2 - x) }

integration gives

(1/8) ln I x + 2 I - (1/2)( 1/ (x+2) ) - (1/8) ln I 2 - x I + C

=> 1/8 [ 4/(x+2) + ln I x + 2 I - ln I 2 - x I ] + C

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If you're just going to use Wolfram, it'll tell you how to decompose fractions. Just get rid of the integral sign:

http://www.wolframalpha.com/input/?i=2%2…

Click on "show steps" in the "partial fraction expansion" part. Note that in the integral steps, it preemptively factored a "2" out of everything for whatever reason.

Done!

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The partial fractions would be
2/ (x+2)^2 * (2-x) = A / (x+2) + B / (x+2)^2 + C / (2-x)
2 = A*(x+2)(2-x) + B*(2-x) + C*(x+2)^2
When x = 2,
2 = C*(2+2)^2 2 = 16*C 1/8 = C
When x = -2,
2 = B*(2+2) 1/2 = B
Finally knowing that all of the x^2 terms must add to 0,
-A + C = 0
A = C
A = 1/8
So the final term that you have to integrate is
[(1/8) / (x+2) + (1/2) / (x+2)^2 + (1/8) / (2-x) ]dx
1
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