How do you integrate #1/(x^2+x) dx#?

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Aug 15, 2016

Answer:

#ln|x/(x+1)|+C#

Explanation:

Let #I=int1/(x^2+x)dx#

#:. I=int1/(x(x+1))dx#

#=int{(x+1)-x}/(x(x+1)) dx#

#=int{(x+1)/(x(x+1))-x/(x(x+1))}dx#

#=int{1/x-1/(x+1)}dx#

#=ln|x|-ln|x+1|#

#=ln|x/(x+1)|+C#

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