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

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Answer 1 · 2016-08-15T17:51:53

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

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

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

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

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

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

$=ln|x|-ln|x+1|$

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

How do you integrate 1/(x^2+x) dx1/(x^2+x) dx?