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I want to approximately compute integral $$I =int_0^1 dx frac{x(2-x)(1-x)}{(1-x)^2+mu x}$$ assuming that $mu$ is small. I tried

Integrate [(2 - x) (1 - x) x/((1 - x)^2 + A x), {x, 0, 1}]

My Mathematica for some reason fails to do this integral explicitly (which is strange, since it is an integral of a rational function; I guess Mathematica obtains divergent answer after decomposing the fraction, but can see that the integral is convergent in fact), so that I could just approximate the exact result.

On the other hand, the point of $mu x$ term is to "regulate" divergence of this integral (i.e. integral without it, $int_0^1 dx frac{x(2-x)(1-x)}{(1-x)^2}$, is logarithmically divergent on the upper limit). Therefore, the whole effect of this term can be accounted by shifting the upper limit:

$$I approx int_0^{1-mu} dx frac{x(2-x)(1-x)}{(1-x)^2}$$

Now Mathematica can compute this integral easily

Integrate [(2 - x) (1 - x) x/(1 - x)^2, {x, 0, 1 - A}, Assumptions -> A > 0]

giving $I = -frac{1}{2}(1+ln(mu^2)-mu^2)$, which can be approximated as $I approx -frac{1}{2}(1+ln(mu^2))$

I wonder if there is any function, smth like ApproximateIntegral[f[x,s],{x,a,b},{s,0}], which could do this whole manipulation for me.

You could use AsymptoticIntegrate, although I change the $mu x$ term to just $mu$, as the latter version is easier for AsymptoticIntegrate to handle:

AsymptoticIntegrate[(x(2-x)(1-x))/((1-x)^2+μ), {x,0,1}, {μ,0,2}]

-1/2 + μ^2/4 + μ (1/2 - Log[μ]/2) - Log[μ]/2

Addendum

AsymptoticIntegrate also works when using $mu x$, but is much slower, and needs to use a higher order than 1 to get a correct answer:

asymp = AsymptoticIntegrate[(x(2-x)(1-x))/((1-x)^2+μ x), {x,0,1}, {μ,0,2}]; //AbsoluteTiming

asymp //TeXForm

{127.263, Null}

$-frac{(mu -1) left(mu ^2-4 mu right) tanh ^{-1}left(frac{mu -2}{sqrt{mu -4}
sqrt{mu }}right)}{2 sqrt{mu -4} sqrt{mu }}+frac{(mu -1) left(mu ^2-4 mu
right) tanh ^{-1}left(frac{sqrt{mu }}{sqrt{mu -4}}right)}{2 sqrt{mu -4}
sqrt{mu }}-mu +frac{1}{4} (mu -2) (mu -1) log (mu )+(mu -1) log (mu
)-frac{2 (mu -3) tanh ^{-1}left(frac{mu -2}{sqrt{mu -4} sqrt{mu
}}right)}{sqrt{frac{mu -4}{mu }}}-frac{(mu -3) (mu -2) tanh
^{-1}left(frac{mu -2}{sqrt{mu -4} sqrt{mu }}right)}{2 sqrt{frac{mu -4}{mu
}}}+frac{2 (mu -3) tanh ^{-1}left(frac{sqrt{mu }}{sqrt{mu
-4}}right)}{sqrt{frac{mu -4}{mu }}}+frac{sqrt{mu -4} (mu -3) sqrt{mu }
left(frac{log (mu )}{2}+frac{(mu -2) tanh ^{-1}left(frac{sqrt{mu
}}{sqrt{mu -4}}right)}{sqrt{mu -4} sqrt{mu }}right)}{2 sqrt{frac{mu
-4}{mu }}}-frac{1}{2}$

The Series expansion of asymp reproduces user64494's result:

Series[asymp, {μ, 0, 2}, Assumptions->μ>0] //TeXForm

$left(-frac{log (mu )}{2}-frac{1}{2}right)+frac{pi sqrt{mu }}{4}+mu
left(-frac{log (mu )}{2}-frac{5}{4}right)+frac{25}{32} pi mu ^{3/2}+mu ^2
left(frac{log (mu )}{2}-frac{19}{24}right)+Oleft(mu ^{5/2}right)$

The true answer in version 12 is as follows.

Integrate[(x*(2 - x)*(1 - x))/((1 - x)^2 +[Mu]*x),{x, 0, 1},Assumptions-> [Mu] > 0 && [Mu] < 1]

(1/(2 (-4 + [Mu])^(
3/2)))Sqrt[[Mu]] (4 I Sqrt[-1 + 4/[Mu]] +
7 I Sqrt[(4 - [Mu]) [Mu]] - 2 I Sqrt[(4 - [Mu]) [Mu]^3] +
I (4 Sqrt[-1 + 4/[Mu]] + 3 Sqrt[(4 - [Mu]) [Mu]] -
5 Sqrt[(4 - [Mu]) [Mu]^3] +
Sqrt[-(-4 + [Mu]) [Mu]^5]) Log[[Mu]] + [Mu] (12 -
7 [Mu] + [Mu]^2) Log[1 - I Sqrt[-([Mu]/(-4 + [Mu]))]] -
12 [Mu] Log[1 + I Sqrt[-([Mu]/(-4 + [Mu]))]] +
7 [Mu]^2 Log[1 + I Sqrt[-([Mu]/(-4 + [Mu]))]] - [Mu]^3 Log[
1 + I Sqrt[-([Mu]/(-4 + [Mu]))]] +
4 Log[1 - (I [Mu])/Sqrt[-(-4 + [Mu]) [Mu]]] - [Mu] Log[
1 - (I [Mu])/Sqrt[-(-4 + [Mu]) [Mu]]] -
4 Log[1 + (I [Mu])/Sqrt[-(-4 + [Mu]) [Mu]]] + [Mu] Log[
1 + (I [Mu])/Sqrt[-(-4 + [Mu]) [Mu]]] -
4 Log[(2 I - I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
Sqrt[-(-4 + [Mu]) [Mu]]] -
11 [Mu] Log[(2 I - I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
Sqrt[-(-4 + [Mu]) [Mu]]] +
7 [Mu]^2 Log[(2 I - I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
Sqrt[-(-4 + [Mu]) [Mu]]] - [Mu]^3 Log[(
2 I - I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
Sqrt[-(-4 + [Mu]) [Mu]]] +
4 Log[(-2 I + I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
Sqrt[-(-4 + [Mu]) [Mu]]] +
11 [Mu] Log[(-2 I + I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
Sqrt[-(-4 + [Mu]) [Mu]]] -
7 [Mu]^2 Log[(-2 I + I [Mu] + Sqrt[-(-4 + [Mu]) [Mu]])/
Sqrt[-(-4 + [Mu]) [Mu]]] + [Mu]^3 Log[(-2 I + I [Mu] +
Sqrt[-(-4 + [Mu]) [Mu]])/Sqrt[-(-4 + [Mu]) [Mu]]])

Series[%, {[Mu], 0, 2}, Assumptions -> [Mu] > 0 && [Mu] < 1]

$$ frac{1}{2} (-log (mu )-1)+frac{pi sqrt{mu }}{4}+frac{1}{4} mu (-2 log (mu )-5)+frac{25}{32} pi mu ^{3/2}+frac{1}{24} mu ^2 (12 log (mu )-19)+Oleft(mu ^{5/2}right)$$

Addition. In order to complete the answer, let us consider the asymptotics for negative values of $mu$:

Integrate[(x*(2 - x)*(1 - x))/((1 - x)^2 + [Mu]*x), {x, 0, 1}, 

PrincipalValue -> True, Assumptions -> [Mu] < 0 && [Mu] > -1]

1/2 (-1 - 2 [Mu] + (-1 - [Mu] + [Mu]^2) Log[-[Mu]] +
Sqrt[[Mu]/(-4 + [Mu])] (-1 - 3 [Mu] + [Mu]^2) Log[
1 + Sqrt[[Mu]/(-4 + [Mu])]] +
Sqrt[[Mu]/(-4 + [Mu])] Log[1 + [Mu]/Sqrt[(-4 + [Mu]) [Mu]]] +
3 [Mu] Sqrt[[Mu]/(-4 + [Mu])]
Log[1 + [Mu]/
Sqrt[(-4 + [Mu]) [Mu]]] - [Mu]^2 Sqrt[[Mu]/(-4 + [Mu])]
Log[1 + [Mu]/Sqrt[(-4 + [Mu]) [Mu]]] -
Sqrt[[Mu]/(-4 + [Mu])]
Log[(2 - [Mu] - Sqrt[(-4 + [Mu]) [Mu]])/
Sqrt[(-4 + [Mu]) [Mu]]] -
3 [Mu] Sqrt[[Mu]/(-4 + [Mu])]
Log[(2 - [Mu] - Sqrt[(-4 + [Mu]) [Mu]])/
Sqrt[(-4 + [Mu]) [Mu]]] + [Mu]^2 Sqrt[[Mu]/(-4 + [Mu])]
Log[(2 - [Mu] - Sqrt[(-4 + [Mu]) [Mu]])/
Sqrt[(-4 + [Mu]) [Mu]]] +
Sqrt[[Mu]/(-4 + [Mu])]
Log[(2 - [Mu] + Sqrt[(-4 + [Mu]) [Mu]])/
Sqrt[(-4 + [Mu]) [Mu]]] +
3 [Mu] Sqrt[[Mu]/(-4 + [Mu])]
Log[(2 - [Mu] + Sqrt[(-4 + [Mu]) [Mu]])/
Sqrt[(-4 + [Mu]) [Mu]]] - [Mu]^2 Sqrt[[Mu]/(-4 + [Mu])]
Log[(2 - [Mu] + Sqrt[(-4 + [Mu]) [Mu]])/
Sqrt[(-4 + [Mu]) [Mu]]])

Series[%, {[Mu], 0, 2}, Assumptions -> [Mu] < 0 && [Mu] > -1]

$$frac{1}{2} (-log (-mu )-1)+frac{1}{4} mu (-2 log (-mu )-5)+frac{1}{24} mu ^2 (12 log (-mu )-19)+Oleft(mu ^{5/2}right) $$