I have encountered this alternating series and I want to know how we can prove that   $$\sum_{n=0}^\infty\frac{(-1)^{n+1}n^2}{n^{3}+1}=\frac{1}{3}-\frac{1}{3}\ln(2)+\frac{\pi}{3}\sinh\left(\pi\frac{\sqrt{3}}{2}\right)$$
So I found this on the Wolfram website today:     So I was wondering about how one might be able to (if possible) solve those four problems by hand. Here are the problems, $\LaTeX$ed:    $ \lim_{x \to +\infty} \dfrac {\sqrt{x^3-x^2+3x}}{\sqrt{x^3}-\sqrt{x^2}+\sqrt{3x}} $ $ \displaystyle\sum_{k=1}^{\infty} \dfrac {(-1)^{k+1} k^2}{k^3+1} $  $ \dfrac {\mathrm{d}}{\mathrm{d}u} \left[ \dfrac {u^{n+1}}{(n+1)^2} \cdot \left[ (n+1) \ln u - 1 \right] \right] $  $ \displaystyle\int_0^{2\pi}\displaystyle\int_0^{\frac{\pi}{4}}\displaystyle\int_0^4 \left( \rho \cos \phi \right) \rho^2 \sin \phi \, \mathrm{d}\rho \mathrm{d}\phi \mathrm{d}\theta$     Ideas   The degree of the numerator is $\frac{3}{2}$. The degree of the denominator is $\frac{3}{2}$. The expression is of an indeterminate form, namely $\frac{\infty}{\infty}$, so we use l'Hoptial's Rule.  No ideas, really.  Derivatives are always pretty easy, although this one is a bit bashy. Basically bash Product and Chain Rules, etc.  Integration by Parts bash? The $\mathrm{d}\theta$ part is very trivial. I mean for the $\rho$ and $\phi$.
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My try, using $x = \sec(u)$ substitution:  $$ \begin{eqnarray} \int \frac{\sqrt{x^2-1}}{x} \mathrm{d}x &amp;=&amp; \int \frac{\sqrt{\sec^2(u) - 1}}{\sec(u)}\tan(u)\sec(u) \mathrm{d}u \\ &amp;=&amp; \int \tan^2(u) \mathrm{d}u \\ &amp;=&amp; \tan(u) - u + C \\ &amp;=&amp; \tan(arcsec(x)) - arcsec(x) + C \end{eqnarray} $$  However, according to , the answer should be: $$ \int \frac{\sqrt{x^2-1}}{x} \mathrm{d}x = \sqrt{x^2-1}+\arctan \left( \frac{1}{\sqrt{x^2-1}} \right)+C $$ When I derive this last answer I don't get back the integrand, but rather: $$ \frac{\mathrm d}{\mathrm d x}\left(\sqrt{x^2-1}+\arctan \left( \frac{1}{\sqrt{x^2-1}} \right)+C\right) = \frac{x}{\sqrt{x^2-1}}- \frac{x}{(x^2-1)^{3/2}\left(1+\frac{1}{x^2-1}\right)} $$  I don't know how to simplify this expression more. Also, I am unable to check whether my answer is correct because I don't know how to find the derivative of $arcsec(x)$.  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