$f(x) = x + x^2$ Let f(x) = f1 (x ) + f2 (x )   where f1 (x ) = x and  f2 (x ) = x2 f1 (x )  is an odd function and f2 (x )  is an even function Hence Fourier expansion of f(x) is sum of Fourier expansion of f1 (x)  and f2 (x). Since f1 (x )  is an odd function $a_0=\ a_n=0\[2ex] \begin{align} \displaystyle a_n&=\frac{2}{\pi{}}\int_0^{\pi{}}f\left(x\right)\sin \ n x\ dx\[2ex] &=\frac{2}{\pi{}}\int_0^{\pi{}}x\sin \ nx\ dx\[2ex] &=\frac{2}{\pi{}}\ {\left[x\left(-\frac{\cos{nx}}{n}\right)-\ \left(1\right)\left(-\frac{\sin{nx}}{n^2}\right)\right]}_0^{\pi{}}\[2ex] &=\frac{2}{\pi{}}\ \left[\left{-\pi{}\frac{{(-1}^n)}{n}-0\right}-\left{0\right}\right]\[2ex] &=\frac{2{(-1)}^{n+1}}{n}\[3ex] {f}_1\left(x\ \right)&=x=2\sum_{n=1}^{\infty{}}\frac{{(-1)}^{n+1}}{n}\sin{nx} \end{align}$ *Since f2 (x)  is an even function* $b_n=0\[2ex] \displaystyle {f}_2\left(x\ \right)=a_0+\ \sum_{n=1}^{\infty{}}\left(a_n\cos{nx}\right)\[3ex] \begin{align} a_0&=\ \frac{1}{\pi{}}\int_0^{\pi{}}f\left(x\right)\ dx\[2ex] &=\frac{1}{\pi{}}\ \int_0^{\pi{}}x^2\ dx\[2ex] &=\frac{1}{\pi{}}{\left[\frac{x^3}{3}\right]}_0^{\pi{}}\[2ex] &=\frac{{\pi{}}^2}{3}\[2ex] \end{align}$ $\begin{align} a_n&=\ \frac{2}{\pi{}}\int_0^{\pi{}}f\left(x\right)\cos{nx}\ dx\[2ex] &=\frac{2}{\pi{}}\int_0^{\pi{}}x^2\cos{nx}\ dx\[2ex] &=\frac{2}{\pi{}}\ {\left[x^2\left(\frac{\sin{nx}}{n}\right)-\ \left(2x\right)\left(-\frac{\cos{nx}}{n^2}\right)\ +(2)\left(-\frac{\sin{nx}}{n^3}\right)\right]}_0^{\pi{}}\[2ex] &=\frac{2}{\pi{}}\left[0+\ 2\pi{}\frac{\cos{nx}}{n^2}-0-0\right]\[2ex] &= 4\frac{{(-1)}^n}{n^2}\[2ex] \end{align}$ $\displaystyle \therefore{} {\ f}_2\left(x\ \right)=\ x^2= \frac{{\pi{}}^2}{3}+4\sum_{n=1}^{\infty{}}\frac{{(-1)}^n}{n^2}\cos{nx}\[2ex] \displaystyle \therefore{}f(x) = x + x^2 = \frac{{\pi{}}^2}{3}+\ 4\sum_{n=1}^{\infty{}}\frac{{(-1)}^n}{n^2}\cos{nx} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +2\sum_{n=1}^{\infty{}}\frac{{(-1)}^{n+1}}{n}\sin{nx}$ ***Now consider*** $\displaystyle x^2 = \frac{{\pi{}}^2}{3}+\ 4\sum_{n=1}^{\infty{}}\frac{{(-1)}^n}{n^2}\cos{nx}\[2ex] Putting \ x = \pi{}\ \ in \ \ (1)$ $\displaystyle \pi{}^2= \frac{{\pi{}}^2}{3}+4\left[-\frac{1}{1^2}\cos{\pi{}+\ \frac{1}{2^2}\cos2{\pi{}-\frac{1}{3^2}\cos{3\pi{}+…}\ }}\right]\[3ex] \displaystyle\frac{{2\pi{}}^2}{3}

= 4\ \left[\frac{1}{1^2}\ +\frac{1}{2^2}\ +\frac{1}{3^2}\ +\frac{1}{4^2}\ +....\ \right]\[3ex] \dfrac{{\pi{}}^2}{6}

Putting x = \ 0 \ in \ \ \ (1)\[3ex] \displaystyle 0 = \frac{{\pi{}}^2}{3} + 4\left[-\frac{1}{1^2}\ +\frac{1}{2^2}-\frac{1}{3^2}\ +\frac{1}{4^2}-....\ \right]\[3ex] \dfrac{{\pi{}}^2}{12}

\ \ Adding \ (2) \ and \ (3)\[4ex] \displaystyle \frac{{\pi{}}^2}{6}+\frac{{\pi{}}^2}{12}=2\left(\frac{1}{1^2}\ +\ \frac{1}{3^2}\ +\frac{1}{5^2}\ +\frac{1}{7^2}\ +....\ \right)\[3ex] \displaystyle \frac{{\pi{}}^2}{8}

= \frac{1}{1^2}\ +\ \frac{1}{3^2}\ +\frac{1}{5^2}\ +\frac{1}{7^2}\ +....\ $