anonymous
  • anonymous
is there anyway to integrate this without using by parts? \int (4x)/(x+1)^2
Mathematics
katieb
  • katieb
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amistre64
  • amistre64
maybe do a partial frac decomp?
amistre64
  • amistre64
or expand the denom and see what its derivative is
amistre64
  • amistre64
x^2 +2x +1 2x + 2 4x +2 -2 = 2(2x+2-2) might be an idea

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amistre64
  • amistre64
well, +4-4 ... but same idea
anonymous
  • anonymous
I did expan the denom, partial fraction will not work because both fractions will have x+1 as the denom
phi
  • phi
you could use u= (x+1)
anonymous
  • anonymous
but du = 1, our number is 4x
anonymous
  • anonymous
*numerator
amistre64
  • amistre64
\[\frac{4x}{x^2+2x+1}=\frac{2(2x+2-2)}{x^2+2x+1}=2*\frac{2x+2}{(x+1)^2}*\frac{-2}{(x+1)^2}\]
phi
  • phi
(4x)/(x+1)^2 dx --> 4(u-1) u^-2 du 4u^(-1) - 4u^(-2) du
amistre64
  • amistre64
umm, mine should have addition in it not * lol
amistre64
  • amistre64
\[\frac{4x}{x^2+2x+1}=\frac{2(2x+2-2)}{x^2+2x+1}=2 \left( \frac{2x+2}{(x+1)^2}+\frac{-2}{(x+1)^2}\right)\]
amistre64
  • amistre64
soo, if we integrate :) 2( ln (x+1)^2 +2(x+1)^-1 +C)
amistre64
  • amistre64
maybe
anonymous
  • anonymous
ok, thx amistre....got it. thanks!!!
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=derivative+4ln%28x%2B1%29+%2B+4%2F%28x%2B1%29 the wolf likes it :)
anonymous
  • anonymous
i didn't play around with the algebra enough. much appreciated.

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