No, it makes no difference: your expectation will just be a function of $\theta$ and $k$, viz.
$$ E[X] = \int_{-\infty}^{\infty} x f(x;k;\theta) \, dx = \int_{\theta}^{\infty} \frac{k\theta^k}{x^k} \, dx = \frac{k\theta}{k-1}, $$
provided $k>1$. (In other words, the other two arguments to $f$ are just parameters: $x$ is the thing in the probability space you're integrating over.)
For the first question regarding integration, let $u = y$ and $dv = n(y- \theta)^{n-1}\, dy.$ It follows that $du = dy$ and $v = (y - \theta)^n$, and integrating by parts we obtain
$$\int ny(y - \theta)^{n-1} \, dy = y(y- \theta)^n - \int(y - \theta)^n \, dy \\ = y(y- \theta)^n - \frac{(y - \theta)^{n+1}}{n+1} + C.$$
Using this anti-derivative, the expected value of the nth order statistic is
$$\begin{align}E(y_{(n)}) &= \int_\theta^{\theta+1} ny (y- \theta)^{n-1} \, dy \\ &= \left. \left[y(y- \theta)^n - \frac{(y - \theta)^{n+1}}{n+1}\right] \right|_\theta^{\theta + 1} \\ &= \theta + \frac{n}{n+1}\end{align} $$
The second moment is
$$\begin{align} E(y_{(n)}^2) &= \int_\theta^{\theta+1} ny^2 (y- \theta)^{n-1} \, dy \end{align}.$$
A first integration by parts yields
$$\begin{align} E(y_{(n)}^2) &= \left.y^2(y- \theta)^n\right|_\theta^{\theta+1} - \int_\theta^{\theta+1} 2y (y- \theta)^{n} \, dy \\ &= (\theta+1)^2 - \int_\theta^{\theta+1} 2y (y- \theta)^{n} \, dy \end{align}$$
A second integration by parts yields
$$\begin{align} E(y_{(n)}^2) &= (\theta+1)^2 - \left. \frac{2y(y - \theta)^{n+1}}{n+1} \right|_\theta^{\theta + 1} + \frac{2}{n+1}\int_\theta^{\theta+1} (y- \theta)^{n+1} \, dy \\ &= (\theta+1)^2 - \frac{2(\theta+1)}{n+1} + \frac{2}{(n+1)(n+2)}\end{align}$$
The variance of the nth order statistic is then
$$E(y_{(n)}^2) - E(y_{(n)})^2 = \frac{n}{(n+1)^2} \underset{n \to \infty}{\longrightarrow} 0$$
Best Answer
First, let $x=\frac{y}{\theta}$ so your integral simplifies to $$ \theta\int_0^1xn(1-x)^{n-1}dx. $$ Now, let $u(x)=x$ and $v(x)=-(1-x)^n$, then integration by parts yields $$ \theta\int_0^1xn(1-x)^{n-1}dx=\theta\int_0^1 u v'=\theta\left(uv|_0^1-\int_0^1 u'v\right)\\ =\theta\left([-x(1-x)^n]|_0^1+\int_0^1(1-x)^ndx\right)=\theta\left(0+\frac{1}{n+1}\right)=\frac{\theta}{n+1}. $$