Integral of -x/(x+1)?

Question

integral of
-x
-----
x + 1
And how do I get it

Answer 1

adjust the numerator in terms of the denominator and then divide term by term

integral of {1-(x+1)} / (x+1)

=integral of { [1/(x+1) ] - 1 }

= ln(x+1) -x + C

or you can divide the numerator by the denominator using long division to get the same breakup with quotient -1 and remainder 1

Answer 2

x / (x + 1) = 1 - 1 / (x + 1)
- x / (x + 1) = 1 / (x + 1) - 1
I = ∫1 / (x + 1).dx - ∫1 .dx
I = log (x + 1) - x + C

Answer 3

∫-x/(x+1) dx
= ∫(-x-1+1)/(x+1) dx
= ∫(-1+1/(x+1) dx
= -x+ln|x+1|+c