How to integrate $\frac{1}{(x+1)(x+2)^2(x+3)^3}$? I tried to solve it with partial fraction decomposition but the expression becomes way too difficult to solve. I could only solve three of six(A-F) expressions of the partial fraction expansion.

Following @paulinho's suggestion, we want to write $\frac{1}{u^2(u-1)(u+1)^3}$ as a sum of partial fractions. The rest we can build by repeatedly using how to write the reciprocal of a quadratic with partial fractions. Note that $$\frac{1}{(u-1)(u+1)}=\frac12\left(\frac{1}{u-1}-\frac{1}{u+1}\right)\\\implies\frac{1}{(u-1)(u+1)^2}=\frac14\left(\frac{1}{u-1}-\frac{1}{u+1}-\frac{2}{(u+1)^2}\right)\\\implies\frac{1}{(u-1)(u+1)^3}=\frac18\left(\frac{1}{u-1}-\frac{1}{u+1}-\frac{2}{(u+1)^2}-\frac{4}{(u+1)^3}\right)\\\implies\frac{1}{u^2(u-1)(u+1)^3}=\frac18\left(\frac{1}{u^2(u-1)}-\frac{1}{u^2(u+1)}-\frac{2}{u^2(u+1)^2}-\frac{4}{u^2(u+1)^3}\right).$$You can do the rest yourself with such observations as$$\frac{1}{u^2(u\pm 1)}=\pm\frac{1}{u}\left(\frac{1}{u}-\frac{1}{u\pm 1}\right)=\pm\frac{1}{u^2}\mp\frac{1}{u}\pm\frac{1}{u\pm 1},\\\frac{1}{u^2(u+1)^2}=\frac{1}{u^2}-\frac{2}{u}+\frac{2}{u+1}+\frac{1}{(u+1)^2}.$$
