<s>[INST] How find $a,b$ if $\int_{0}^{1}\frac{x^{n-1}}{1+x}dx=\frac{a}{n}+\frac{b}{n^2}+o(\frac{1}{n^2}),n\to \infty$ let
$$\int_{0}^{1}\dfrac{x^{n-1}}{1+x}dx=\dfrac{a}{n}+\dfrac{b}{n^2}+o(\dfrac{1}{n^2}),n\to \infty$$
Find the $a,b$
$$\dfrac{x^{n-1}}{1+x}=x^{n-1}(1-x+x^2-x^3+\cdots)=x^{n-1}-x^n+\cdots$$
so
$$\int_{0}^{1}\dfrac{x^{n-1}}{1+x}=\dfrac{1}{n}-\dfrac{1}{n+1}+\dfrac{1}{n+2}-\cdots$$
and note
$$\dfrac{1}{n+1}=\dfrac{1}{n}\left(\frac{1}{1+\dfrac{1}{n}}\right)=\dfrac{1}{n}-\dfrac{1}{n^2}+\dfrac{1}{n^3}+o(1/n^3)$$
and simaler
$$\dfrac{1}{n+2}=\dfrac{1}{n}-\dfrac{2}{n^2}+o(1/n^2)$$
$$\dfrac{1}{n+3}=\dfrac{1}{n}-\dfrac{3}{n^2}+o(1/n^2)$$
then which term end?
 [/INST] Here's a possible to find the coefficients:
From 
$$
n\int_0^1 \frac{x^{n-1}}{1+x} d\;x =\left[\frac{x^n}{1+x} \right]^1_0 + \int_0^1 \frac{x^n}{(1+x)^2} d\;x =1/2 + \int_0^1 \frac{x^n}{(1+x)^2} d\;x 
$$
Since the last term tends to zero as $n$ increases, we get $a =1/2$.
The same process allows to determine $b$.
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