Evaluate $\int x^2\log(1-x^2)dx$, and hence prove that $\frac1{1\cdot5}+\frac1{2\cdot7}+\frac1{3\cdot9}+...=\frac23\log2-\frac89$
My Attempt:
Integrating by parts,
$$\log(1-x^2)\cdot\frac{x^3}3-\int\frac{-2x}{1-x^2}\cdot\frac{x^3}3dx\\=\frac{x^3}3\log(1-x^2)-\frac23\int\frac{1-x^4-1}{1-x^2}dx\\=\frac{x^3}3\log(1-x^2)-\frac23\int1+x^2-\frac1{1-x^2}dx\\=\frac{x^3}3(\log(1-x)+\log(1+x))-\frac23(x+\frac{x^3}3)+\frac13\log|\frac{1-x}{1+x}|+c\\=\frac{x^3-1}3\log(1-x)+\frac{x^3+1}3\log(1+x)-\frac23(x+\frac{x^3}3)+c$$
If we apply limits from $0$ to $1$, we get
$$\frac23\log2-\frac89,$$
which is the required RHS.
How to get the required LHS?
Using the Taylor series of $\log(1-t)$: $$\log(1-t)=-\sum_{n=1}^\infty\frac{t^n}n,\qquad t\in[0,1),$$ we have $$\int_0^1 x^2\log(1-x^2)dx=-\int_0^1x^2\sum_{n=1}^\infty\frac{x^{2n}}n\,dx=-\sum_{n=1}^\infty\frac1n\int_0^1x^{2+2n}\,dx=-\sum_{n=1}^\infty\frac1{n(3+2n)},$$ which is exactly the required LHS (except for the minus sign). Here, the reason why we can interchange the summation and integration is the Tonelli's theorem, or that we can always integration a power series term by term, see here.
So, we have proven that $$-\sum_{n=1}^\infty\frac1{n(3+2n)}=\frac23\log2-\frac89,$$ which differs from the original problem by a minus sign. Indeed, the original problem is not correct. Clearly we have $\frac1{1\cdot5}+\frac1{2\cdot7}+\frac1{3\cdot9}+\cdots>0$, but $\frac23\log2-\frac89< \frac23-\frac89<0.$ Therefore, the right version is $$\frac1{1\cdot5}+\frac1{2\cdot7}+\frac1{3\cdot9}+\cdots=\frac89-\frac23\log2.$$ Wlofram alpha agrees with our computation.