Asymptotics of a function I have to find the asymptotics of the following integral
$$\int^1_0 \frac{\sin(x)}{x(1+x)^n} dx$$
as $n\to\infty$.
I know I am supposed to use the lebesgue dominated convergence theorem and create a sequence $b_n$ and aim for $\lim_{n\to\infty} a_n/b_n =1$ but I can't seem to get my head round it.
Any help I would be very much grateful for
 A: Denote the integral in question by $I_n$ and squeeze it baby!
Notice that the inequalities $x-x^3/3!<\sin(x)<x$ hold on $x\in(0,1)$
so
$$
\int_0^1dx\frac{1-x^2/3!}{(1+x)^n}<I_n<
\int_0^1dx\frac{1}{(1+x)^n}
$$
doing the algebra (which is elementary) you get
$$
\frac{1}{n-1}+\mathcal{O}(n^{-2})<I_n<\frac{1}{n-1}+\mathcal{O}( n^{-2})
$$
which shows that $I_n\sim n^{-1}$ or $\lim_{n\rightarrow\infty}nI_n=1$

For the record: 
Integration by parts also works nicely here: use $u(x)=\text{sinc}(x),\,v'(x)=(1+x)^{-n}$
A: Let $I_n$ be given by
$$\begin{align}
I_n&=\int_0^1 \frac{\sin(x)}{x(1+x)^n}\,dx\\\\
&=\int_0^1 (1+x)^{-n}\,dx+\sum_{k=1}^\infty\frac{(-1)^{k}}{(2k+1)!}\int_0^1x^{2k}(1+x)^{-n}\,dx\\\\
&=\frac{1}{n-1}-\frac{2}{2^n(n-1)}+\sum_{k=1}^\infty\frac{(-1)^{k}}{(2k+1)!}\int_0^1 x^{2k}(1+x)^{-n}\,dx\tag 1
\end{align}$$
We can evaluate the integral on the right-hand side of $(1)$ be making the substitution $x\to x-1$ and using the binomial theorem to write $(x-1)^{2k}=\sum_{\ell = 0}^{2k}\binom{2k}{\ell}(-1)^{2k-\ell}x^\ell$.  Proceeding we find
$$\int_0^1x^{2k}(1+x)^{-n}\,dx=\sum_{\ell = 0}^{2k}\binom{2k}{\ell}(-1)^{2k-\ell}\frac{1-2^{-n+\ell+1}}{n-(\ell+1)}$$
which shows that $\int_0^1x^{2k}(1+x)^{-n}\,dx=O\left(\frac{1}{n^{2k+1}}\right)$
Finally, we have 
$$I_n=\frac{1}{n-1}-\frac{1}{3(n-1)(n-2)(n-3)}+O\left(\frac1{n^5}\right)$$
