How do i solve this integral?

i don't know what to do with the m
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1 Answer
George C.
Mar 5, 2018

#int dx/(x(x+1)(x+2)...(x+m))= sum_(k=0)^m ((-1)^k ln abs(x+k))/(k!(m-k)!) + C#

Explanation:

Writing:

#1/(x(x+1)(x+2)...(x+m)) = a_0/x + a_1/(x+1) +a_2/(x+2)+...+a_m/(x+m)#

Then using Oliver Heaviside's cover up method, we find:

#a_0 = 1/((color(blue)(0)+1)((color(blue)(0)+2)...((color(blue)(0)+m))#

#a_1 = 1/((color(blue)(-1))((color(blue)(-1))+2)((color(blue)(-1))+3)...((color(blue)(-1))+m)#

#vdots#

#a_k = 1/(prod_(j=0, j!=k)^m(-k+j)) = (-1)^k/(k!(m-k)!)#

#vdots#

#a_m = 1/((color(blue)(-m))((color(blue)(-m))+1)((color(blue)(-m))+2)...((color(blue)(-m))+m-1))#

Then:

#int dx/(x(x+1)(x+2)...(x+m))#

#=int sum_(k=0)^m a_k/(x+k) dx#

#= sum_(k=0)^m a_k ln abs(x+k) + C#

#= sum_(k=0)^m ((-1)^k ln abs(x+k))/(k!(m-k)!) + C#

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