<s>[INST] Finding $f(2019)$ in definite integration If $\displaystyle f(n)=\int^{1}_{0}(1+x+x^2+\cdots +x^{n-1})(1+3x+5x^2+\cdots +(2n-1)x^{n-1})dx$. Then $f(2019)$ is 
What I tried:
$$1+x+x^2+\cdots +x^{n-1}=\frac{1-x^n}{1-x}$$
and $$1+3x+5x^2+\cdots +(2n-1)x^{n-1}=\frac{1}{1-x}+\frac{2(1-x^n)}{1-x}-\frac{(2n-1)x^n}{1-x}$$
How do I solve it? Help me, please.
 [/INST] By making change of variable $x=t^2$, we obtain
$$\begin{align*}
f(n)&=2\int_0^1 t(1+t^2+\cdots +t^{2n-2})(1+3t^2+5t^4+\cdots +(2n-1)t^{2n-2})dt.
\end{align*}$$ Further making substitution $u=t+t^3+\cdots +t^{2n-1}$, we get
$$
f(n)=2\int_0^n u\ du=\left[u^2\right]^n_0=n^2.
$$ So $f(2019)=2019^2$ follows.
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