Fourier Series Coefficients Calculation
I try it the long way to give a proof of your simplified answer, and from the very first step to make sure that everybody knows what we are talking about (and to be very careful with the signs).
Problem:
$$\int_{-1}^1te^{-jn \pi\ t}\ dt  =?=?=  \frac{e^{jn \pi\ }  + e^{-jn \pi\ }}{-jn \pi\ } $$

Let´s do it with the integration by parts. The template:

$$\int_{-a}^bf'(x)g(x)dx=f(b)g(b)-f(a)g(a)-\int_{a}^bf(x)g'(x)dx$$

We choose the following substitutions:
$$f'(x)=e^{-jn \pi\ t}$$ which integrates to 
$$f(x)=-\frac{e^{-jn \pi\ t}}{jn \pi\ t}$$
and
$$g(x) = t$$
derivative: $$g'(x) = 1$$

Next step, fill out, which leads to :

$$-\frac{e^{-jn \pi\ }}{jn \pi\ }(1) - \frac{e^{jn \pi\ } }{jn \pi\ }(1) - \int_{a}^b- \frac{e^{-jn \pi\ t}}{jn \pi\ t}(1)dt$$

(the first two (1)´s are the values for upper and lower boundary of the integral, the third (1) is the derivative g'(x), of $$g(x)$$, which is 1)

which, if further dissolved, leads to (consider the signs of the boundaries)

$$-\frac{e^{-jn \pi\ }}{jn \pi\ } - \frac{e^{jn \pi\ } }{jn \pi\ } - (-\frac{e^{jn \pi\ }}{jn \pi\ } - \frac{e^{jn \pi\ }}{-jn \pi\ } ) $$

and, only solving the term within the braces (how can we scratch terms out?):

$$-\frac{e^{-jn \pi\ }}{jn \pi\ } - \frac{e^{jn \pi\ } }{jn \pi\ } +\frac{e^{jn \pi\ }}{jn \pi\ } - \frac{e^{jn \pi\ }}{jn \pi\ } $$

which leads at the end to:

$$\frac{e^{jn \pi\ }  + e^{-jn \pi\ }}{-jn \pi\ } $$

which is the proof for your assumption.

regards