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 March 19th, 2011, 10:58 PM #1 Member   Joined: Aug 2009 Posts: 41 Thanks: 0 crazy partial fraction integral integral of (x)dx/((3x+7)^3) i have no idea how to solve for this integral , i'm learning integral by partial fraction and this is the hardest one that i can't do , if u know how to solve , please help me , thank you so much
 March 20th, 2011, 03:01 AM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: crazy partial fraction integral $\int\frac{x}{(3x+7)^3}\mathrm{d}x$ Don't use partial fractions for this one, just set u = 3x+7, so du = 3 dx and x = (u-7)/3. $\frac13\int\frac{\frac13u-\frac73}{u^3}\mathrm{d}u \\= \frac19\int u^{-2}-7u^{-3}\mathrm{d}u\\ =\frac19\left[\frac{1}{-1}u^{-1}-\frac{7}{-2}u^{-2}\right]+c\\ =-\frac{1}{9u}+\frac{7}{18u^2}+c\\ =-\frac{1}{9(3x+7)}+\frac{7}{18(3x+7)^2}+c$
 March 25th, 2011, 07:53 PM #3 Member   Joined: Aug 2009 Posts: 41 Thanks: 0 Re: crazy partial fraction integral thank you so much, why didn't i think about using u substitution !!!

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