Does $\lim_{n \to +\infty} \frac{1}{n^2} \sum_{k=1}^{n} k \ln\left( \frac{k^2+n^2}{n^2}\right )$ exist? 
I need to examine whether the following limit exists, or not.
  $$\lim_{n \to +\infty} \frac{1}{n^2} \sum_{k=1}^{n} k \ln\left( \frac{k^2+n^2}{n^2}\right )$$
   If it does, I need to calculate its value.  

How to even start this? I've got no idea.
 A: Just to complete Clement C.'s answer, integration by parts leads to:
$$\begin{eqnarray*} I=\int_{0}^{1}x \log(1+x^2)\,dx &=& \left.\frac{x^2}{2}\log(1+x^2)\right|_{0}^{1}-\int_{0}^{1}\frac{x^3}{1+x^2}\,dx\\&=&\frac{\log(2)}{2}-\int_{0}^{1}x\,dx+\int_{0}^{1}\frac{x\,dx}{1+x^2} \\&=&\color{red}{\log(2)-\frac{1}{2}}.\end{eqnarray*}$$
Another chance is given by termwise integration of a Taylor series:
$$\begin{eqnarray*}I = \int_{0}^{1}\sum_{n\geq 1}\frac{(-1)^{n+1}x^{2n+1}}{n}\,dx&=&\frac{1}{2}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n(n+1)}\\&=&\frac{1}{2}\left(\sum_{n\geq 1}\frac{(-1)^{n+1}}{n}-\sum_{n\geq 1}\frac{(-1)^{n+1}}{n+1}\right)\\&=&\frac{1}{2}\left(2\sum_{n\geq 1}\frac{(-1)^{n+1}}{n}-1\right)=\color{red}{\log(2)-\frac{1}{2}}.\end{eqnarray*}$$
A third way is given by the substitution $x=\sqrt{z-1}$ plus Feynman's trick, from which:
$$\begin{eqnarray*} I = \frac{1}{2}\int_{1}^{2}\log(z)\,dz = \left.\frac{1}{2}\frac{d}{d\alpha}\int_{1}^{2}z^{\alpha}\,dz\,\right|_{\alpha=0}&=&\left.\frac{1}{2}\frac{d}{d\alpha}\frac{2^{\alpha}-1}{\alpha}\,\right|_{\alpha=1}\\&=&\left.\frac{1+2^a(a\log 2-1)}{2a^2}\right|_{\alpha=1}\\&=&\color{red}{\log(2)-\frac{1}{2}}.\end{eqnarray*}$$
A: An idea: "Riemann sums" may be a good start.
Massage your current sum into something of the form
$$
\frac{1}{n}\sum_{k=0}^n \frac{k}{n} \ln \left( 1+\left(\frac{k}{n}\right)^2\right)
$$
and recognize a Riemann sum for the (continuous) function $f\colon[0,1]\to\mathbb{R}$ defined by $f(x) = x\ln(1+x^2)$.

Update: Jack d'Aurizio gave a way (actually, two) to evaluate the integral $$\int_0^1 x\ln(1+x^2)dx$$ in his separate answer, which complements this one.
A: Another approach. Using Abel's summation we have $$S=\sum_{k=0}^{n}k\log\left(1+\left(\frac{k}{n}\right)^{2}\right)=\frac{n\left(n+1\right)\log\left(2\right)}{2}-\int_{0}^{n}\frac{\left\lfloor t\left(t+1\right)\right\rfloor t}{n^{2}+t^{2}}dt
 $$ where $\left\lfloor x\right\rfloor 
 $ is the floor function. Since $\left\lfloor x\right\rfloor =x+O\left(1\right)
 $ we have $$S=\frac{n\left(n+1\right)\log\left(2\right)}{2}-\int_{0}^{n}\frac{t^{2}\left(t+1\right)}{n^{2}+t^{2}}dt+O\left(1\right)
 $$ and the integral is not too complicated $$\int_{0}^{n}\frac{t^{2}\left(t+1\right)}{n^{2}+t^{2}}dt=-n^{2}\int_{0}^{n}\frac{t}{n^{2}+t^{2}}dt-n\int_{0}^{n}\frac{1}{n^{2}+t^{2}}dt+\int_{0}^{n}tdt+\int_{0}^{n}1dt
 $$ $$=-\frac{1}{4}n^{2}\log\left(4\right)+\frac{n^{2}}{2}-\frac{\pi}{4}n+n
  $$ hence $$\frac{1}{n^{2}}\sum_{k=0}^{n}k\log\left(1+\left(\frac{k}{n}\right)^{2}\right)=\log\left(2\right)+\frac{\log\left(2\right)}{n}-\frac{1}{2}-\frac{\pi}{4n}+\frac{1}{n}+O\left(\frac{1}{n}\right)\rightarrow\log\left(2\right)-\frac{1}{2}
 $$ as $n\rightarrow\infty$.
