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Evaluation of $\displaystyle \int\frac{1}{(1+x)(2+x)^2(3+x)^3}dx$

$\bf{My\; Trial \; Solution::}$ Given $\displaystyle \int\frac{1}{(1+x)(2+x)^2(3+x)^3}dx\;\;\;,$ Let $(3+x)=y\;,$ Then $dx = dy$

So Integral Convert into $\displaystyle \int\frac{1}{(y-2)(y-1)^2y^3}dy\;\;,$ Now Let $\displaystyle y = \frac{1}{z}\;\;,$ Then $\displaystyle dy = -\frac{1}{z^2}dz$

So Integral convert into $\displaystyle \int\frac{z^6}{(2z-1)(z-1)^2}\cdot \frac{1}{z^2}dz = \int\frac{z^4}{(2z-1)(z-1)^2}dz$

Now How can I solve after that

Help Required

Thanks

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    $\begingroup$ I think that you have to use partial fraction decomposition of the initial rational function. $\endgroup$ – kmitov Feb 15 '14 at 15:42
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    $\begingroup$ look at : en.wikipedia.org/wiki/Partial_fraction_decomposition $\endgroup$ – user119228 Feb 15 '14 at 15:42
  • $\begingroup$ Partial fraction decomposition is the ONLY way. $\endgroup$ – Claude Leibovici Feb 15 '14 at 16:18
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Use partial fraction decomposition to decompose the expression into (I used the Wolf): $$\frac{1}{(x+1) (x+2)^2 (x+3)^3} = \frac{2}{x+2}-\frac{17}{8}\frac{1}{x+3}-\frac{1}{(x+2)^2}-\frac{5}{4}\frac{1}{(x+3)^2}-\frac{1}{2}\frac{1}{(x+3)^3}+ \frac{1}{8}\frac{1}{x+1}$$

and then integrate directly.

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$\frac{1}{(1+x)(2+x)^2(3+x)^3}=\frac{A}{1+x}+\frac{B}{2+x}+\frac{C}{(2+x)^2}+\frac{D}{3+x}+\frac{E}{(3+x)^2}+\frac{F}{(3+x)^3}$

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