but can't figure out HOW they derived the partial fractions (the 3rd step)?????
Can anyone please explain?
Here's the link:
http://www.wolframalpha.com/input/?i=int…
incase the link doesn't work, my original problem was: integral of 2/ (x+2)^2(2x) dx
Can anyone please explain?
Here's the link:
http://www.wolframalpha.com/input/?i=int…
incase the link doesn't work, my original problem was: integral of 2/ (x+2)^2(2x) dx

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
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

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!
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!

The partial fractions would be
2/ (x+2)^2 * (2x) = A / (x+2) + B / (x+2)^2 + C / (2x)
2 = A*(x+2)(2x) + B*(2x) + 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) / (2x) ]dx
2/ (x+2)^2 * (2x) = A / (x+2) + B / (x+2)^2 + C / (2x)
2 = A*(x+2)(2x) + B*(2x) + 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) / (2x) ]dx