代码：
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代码:
\[
\begin{aligned}
\int \frac{\frac{1}{x^2}}{(\cosh \frac{1}{x})^2}\, dx &= 4\int \frac{\frac{1}{x^2}}{(e^{\frac{1}{x}} + e^{-\frac{1}{x}})^2}\, dx
\\
&= 4\int \frac{\frac{1}{x^2}}{e^{\frac{2}{x}} + e^{-\frac{2}{x}} + 2}\,dx \qquad (\mbox{let } e^{\frac{2}{x}} = u, \mbox{ then } x = \frac{2}{\ln u})
\\
&= 4\int \frac{\frac{(\ln u)^2}{4}}{u + u^{-1}+ 2}(-2)\frac{1}{(\ln u)^2}\frac{1}{u} \, du \qquad (dx = -2\frac{1}{(\ln u)^2}\frac{1}{u} du)
\\
&= -2\int \frac{1}{(u+1)^2} \,du
\\
&= \frac{1}{u+1} + C
\\
&= \frac{1}{e^{\frac{2}{x}}+1} + C
\end{aligned}
\]
Similarly we could obtain:
\[
\int \frac{\frac{1}{x^2}}{(\sinh \frac{1}{x})^2}\, dx = \frac{1}{e^{\frac{2}{x}}-1} + C
\]
Therefore:
\[
\begin{aligned}
\int B \left[\frac{\frac{C}{T}}{\sinh(\frac{C}{T})} \right]^2 \,dT &= \frac{BC}{e^{\frac{2C}{T}} - 1} + c_1
\\
\int D \left[\frac{\frac{E}{T}}{\cosh(\frac{E}{T})} \right]^2 \,dT &= \frac{DE}{e^{\frac{2E}{T}} + 1} + c_2
\end{aligned}
\]
And $y$ is as follows:
\[
y = AT + \frac{BC}{e^{\frac{2C}{T}} - 1} + \frac{DE}{e^{\frac{2E}{T}} + 1} + c
\]
where $c$ is a generic constant.编译见附件：
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