I know that $\sum_{n=0}^{\infty}\frac{1}{4n^2-1}$ is a telescoping sum, but $x^{2n+1}$ in the sum complicates it a bit.
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The sum is $-x+\dfrac {x^3}3+\dfrac{x^5}{15}+\dfrac{x^7}{35}+\dfrac{x^9}{63}+\cdots+\dfrac{x^{2n+1}}{4n^2-1}+\cdots$.
Its derivative is $-1+x^2+\dfrac{x^4}3+\dfrac{x^6}5+\dfrac{x^8}{7}+\cdots+\dfrac{x^{2n}}{2n-1}+\cdots.$
$=-1+x\left(x+\dfrac{x^3}3+\dfrac{x^5}5+\dfrac {x^7}7+\cdots+\dfrac{x^{2n-1}}{2n-1}+\cdots\right)$
$=-1+x\tanh^{-1}x$.
Can you take it from here?
J. W. Tanner
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Taking $\int_{0}^{X} f'(x) dx$? I haven't really worked with $tanh$ or $arctanh$ before. – Labbsserts Nov 24 '20 at 23:12
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2$\tanh x=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}$, so $\tanh^{-1}x=\frac12\ln(1+x)-\frac12\ln(1-x)$ – J. W. Tanner Nov 24 '20 at 23:33
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