Integral help

cocomo

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Aug 27, 2009
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Integral from [0, 1]: x-1/2x+3 dx
(I've tried splitting it up x/2x+3 - 1/2x+3. I know the second part is a 1/2ln(2+3) but can't figure out how to solve for the first part)

AND

Integral: lnx/(2x-1)^2 dx
(clueless)
 
cocomo said:
Integral from [0, 1]: x-1/2x+3 dx
(I've tried splitting it up x/2x+3 - 1/2x+3. I know the second part is a 1/2ln(2+3) but can't figure out how to solve for the first part
Write it as x12x+3=1254x+6\displaystyle \frac{{x - 1}}{{2x + 3}} = \frac{1}{2} - \frac{5}{{4x + 6}}.
 
x1/2x+3 = x+62\displaystyle x-1/2x+3 \ = \ \frac{x+6}{2}

(x1)/2x+3 = x2x+62\displaystyle (x-1)/2x+3 \ = \ \frac{x^{2}-x+6}{2}

x1/(2x+3) = 2x2+3x12x+3\displaystyle x-1/(2x+3) \ = \ \frac{2x^{2}+3x-1}{2x+3}

(x1)/(2x+3) = x12x+3\displaystyle (x-1)/(2x+3) \ = \ \frac{x-1}{2x+3}

Now, which one are you talking about?\displaystyle Now, \ which \ one \ are \ you \ talking \ about?
 
If, as I believe, the last form (x - 1)/(2x + 3) is what you intended, use long division to get:

1/2 - (5/2) [1/(2x + 3)]

Integrating 1/2 is easy; use a u-sub: u = 2x + 3, to integrate the rational function.
 
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