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How do I find the indefinite integral of this expression? :

$$\int \frac{dx}{x^{3}(2+3x)^2}$$

Thanks in advance!

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HINT:

$I=\int\dfrac{dx}{x^3(3x+2)^2}=\dfrac{27}8\int\dfrac{(3x+2-3x)^3}{(3x)^3(3x+2)^2}dx$

$8I/27=\int\dfrac{(3x+2)^3-3(3x+2)^23x+3(3x+2)(3x)^2-(3x)^3}{(3x)^3(3x+2)^2}dx$

$=\int\dfrac{3x+2}{(3x)^3}dx-\int\dfrac{dx}{3x^2}+\int\dfrac1{3x(3x+2)}dx-\int\dfrac{dx}{(3x+2)^2}$

Finally use $\dfrac1{3x(3x+2)}=\dfrac12\cdot\dfrac{(3x+2)-3x}{3x(3x+2)}$

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Use partial fraction to decompose your integral into a bunch of proper fractions$$\frac 1{x^3(2x+3)^2} = \frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{2x+3}+\frac{E}{(2x+3)^2}$$

You should find the following; $$A=\frac{27}{16}$$ $$B=\frac{-3}{4}$$ $$C=\frac{1}{4}$$ $$D=\frac{-81}{16}$$ $$E=\frac{-27}{8}$$

Now you just integrate all the the fractions(use u substitution on $(2x + 3)^2$ if you have to. You will be left with;

$$\frac{27}{16}\ln|x| + \frac{3}{4x} - \frac{1}{8x^2} - \frac{27}{16}\ln|3x+2| + \frac{9}{8(2+3x)} + C$$

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Evaluate $~I(a,b)=\displaystyle\int\frac{dx}{(x+a)(3x+b)}$ first, by using partial fraction decomposition,

and then express your integral in terms of $I^{^{\Large(3,~2)}}(0,2)$.

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$ \bf hint:$ use partial fraction to decompose $$\frac 1{x^3(2x+3)^2} = \frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{2x+3}+\frac{E}{(2x+3)^2}$$

the integrals $$\int \frac{dx}{2x+3} = \frac 12\ln|2x+3|, \int \frac{dx}{(2x+3)^2} = -\frac{1}{2(2x+3)}$$

in fact what you take away from this is that $\int f(ax+b)\, dx$ is as easy/hard as $\int f(u) \, du.$

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    $\begingroup$ You are too fast for the old man ! This is not fair game being faster than the light !! Cheers $\endgroup$ – Claude Leibovici Mar 14 '15 at 15:07
  • $\begingroup$ But then the last term would be pretty complicated, wouldn't it, with a linear numerator and quadratic denominator? $\endgroup$ – JavaNewbie_M107 Mar 14 '15 at 15:08
  • $\begingroup$ Sorry, my bad. That's for unfactorable quadratics. Got muddled up a bit :). $\endgroup$ – JavaNewbie_M107 Mar 15 '15 at 6:38

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