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

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