Integral of (2x + 3) / (9x^2 - 12x + 8) dx

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March 1, 2013 - 1:58pm

hoya

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Joined: Mar 1 2013 - 1:42pm

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Integral of (2x + 3) / (9x^2 - 12x + 8) dx

Is anyone there can help me with this problem integral of (2x+3)/(9x²-12x+8) dx

Thank you.

March 5, 2013 - 3:54pm

Romel Verterra

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Joined: Oct 12 2008 - 3:39pm

Points: 1846

Re: Integral of (2x + 3) / (9x^2 - 12x + 8) dx

$\displaystyle \int \dfrac{2x + 3}{9x^2 - 12x + 8} ~ dx$

$= \displaystyle \int \dfrac{2x + 3}{9(x^2 - \frac{4}{3}x) + 8} ~ dx$

$= \displaystyle \int \dfrac{(2x - \frac{4}{3}) + \frac{13}{3}}{9(x^2 - \frac{4}{3}x) + 8} ~ dx$

$= \displaystyle \int \dfrac{2x - \frac{4}{3}}{9(x^2 - \frac{4}{3}x) + 8} ~ dx + \int \dfrac{\frac{13}{3}}{9(x^2 - \frac{4}{3}x) + 8} ~ dx$

$= \displaystyle \dfrac{1}{9} \ln [ \, 9(x^2 - \frac{4}{3}x) + 8 \, ] + \dfrac{13}{3}\int \dfrac{dx}{9(x^2 - \frac{4}{3}x + \frac{4}{9}) + 8 - 9(\frac{4}{9})}$

$= \displaystyle \frac{1}{9} \ln (9x^2 - 12x + 8) + \dfrac{13}{3}\int \dfrac{dx}{9(x - \frac{2}{3})^2 + 4}$

$= \displaystyle \frac{1}{9} \ln (9x^2 - 12x + 8) + \dfrac{13}{3}\int \dfrac{dx}{[ \, 3(x - \frac{2}{3}) \, ]^2 + 2^2}$

$= \displaystyle \frac{1}{9} \ln (9x^2 - 12x + 8) + \dfrac{13}{3} \left( \dfrac{1}{3} \right) \int \dfrac{3 ~ dx}{[ \, 3(x - \frac{2}{3}) \, ]^2 + 2^2}$