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\int {\frac  {1}{(x+1)(x+2)(x+3)}}\,dx\,

Again, this problem is a Partial Fractions problem. Let

{\frac  {1}{(x+1)(x+2)(x+3)}}={\frac  {A}{x+1}}+{\frac  {B}{x+2}}+{\frac  {C}{x+3}}\,

Now, multiply through to get common denominators and equate the numerators.


Pick values of x\, to cancel out all but one term.

Let x=-1\,, then 1=2A\, or A={\frac  {1}{2}}\,
Let x=-2\,, then 1=-B\, or B=-1\,
Let x=-3\,, then 1=2C\, or C={\frac  {1}{2}}\,

Thus our integral becomes

\int {\frac  {1}{(x+1)(x+2)(x+3)}}\,dx=\int \left({\frac  {0.5}{x+1}}-{\frac  {1}{x+2}}+{\frac  {0.5}{x+3}}\right)\,dx={\frac  {1}{2}}\ln |x+1|-\ln |x+2|+{\frac  {1}{2}}\ln |x+3|+C\,