The function is $f(x)=\begin{cases}0,&-\pi\leq x \leq 0\\x,&0<x\leq \pi\end{cases}$.
The Fourier series of the function is: $$\frac{1}{2}(\frac{\pi}{2})+\sum_{n=1}^\infty \frac{(-1)^n-1}{\pi n^{2}}\cos (nx)+\frac{(-1)^{n+1}}{n}\sin (nx)$$ Since the function can be integrated termwise, the Fourier series for the antiderivative $\int_{-\pi}^xf(x) \,dx$ is: $$\frac{1}{2}(\frac{\pi}{2})(x+\pi)+\sum_{n=1}^\infty \frac{1}{n}\left( \frac{(-1)^n-1}{\pi n^{2}}\sin (nx)+\frac{(-1)^{n}}{n}(\cos (nx)-(-1)^n)\right)$$ The antiderivative of the function is $F(x)=\begin{cases}0,&-\pi\leq x \leq 0\\\frac{x^2}{2},&0<x\leq \pi\end{cases}$. The Fourier series of the antiderivative is: $$\frac{1}{2}(\frac{\pi^2}{6})+\sum_{n=1}^\infty \frac{1}{n}\left(\left(\frac{(-1)^n-1}{\pi n^{2}}-(-1)^n\frac{\pi}{2}\right)\sin (nx)+\frac{(-1)^{n}}{n}\cos (nx)\right)$$
Where I am missing? Why I am having wrong answer. Even if I did some mistakes in calculations, what's with $x$ term obtained in piece wise integration. Thanks in advance.
From comments, What you have is: $$\frac{\pi}{4}x+\frac{\pi^2}{4}+\sum \frac{1}{n}\left(\frac{(-1)^n-1}{\pi n^2}\sin(nx)+\frac{(-1)^n}{n}\cos(nx)\right)-\sum\frac{1}{n^2}$$ The Fourier expansion of $x$ on $[-\pi, \pi]$ is $x=-2\sum \frac{(-1)^n}{n}\sin (nx)$ and using the fact that $\sum\frac{1}{n^2}=\frac{\pi^2}{6}$. It is obttained that, $$\frac{\pi}{4}\left(-2\sum \frac{(-1)^n}{n}\sin (nx)\right)+\frac{\pi^2}{4}+\sum \frac{1}{n}\left(\frac{(-1)^n-1}{\pi n^2}\sin(nx)+\frac{(-1)^n}{n}\cos(nx)\right)-\frac{\pi^2}{6}$$ $$\frac{\pi^2}{12}+\sum\frac{1}{n} \left(\left(\frac{(-1)^n-1}{\pi n^2}-(-1)^n\frac{\pi}{2}\right)\sin(nx)+\frac{(-1)^n}{n}\cos(nx)\right)$$ Now both results conicides.