Integral help

[cocomo]

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)

 

[pka]

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 \(\displaystyle \frac{{x - 1}}{{2x + 3}} = \frac{1}{2} - \frac{5}{{4x + 6}}\).

 

[BigGlenntheHeavy]

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

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

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

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

\(\displaystyle Now, \ which \ one \ are \ you \ talking \ about?\)

 

[fasteddie65]

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.