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**Hint:**These types of integral problems are pretty straight forward and are very easy to solve. In the given problem we need to find the integral for a hyperbolic function. First of all we need all the general formulae and equations for hyperbolic functions. Using the general formulae and putting them in the integral, we can easily evaluate the answer. The integral for the hyperbolic functions are similar to that of the general trigonometric functions. Some of the very basic and general equations for hyperbolic functions include,

\[\begin{align}

& \sinh x=\dfrac{{{e}^{x}}-{{e}^{-x}}}{2} \\

& \cosh x=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2} \\

& \tanh x=\dfrac{\sinh x}{\cosh x} \\

& \coth x=\dfrac{\cosh x}{\sinh x} \\

& \operatorname{sech}x=\dfrac{1}{\cosh x} \\

& \text{cosechx=}\dfrac{1}{\sinh x} \\

\end{align}\]

**Complete step by step answer:**

Now, we start off with our given solution, we write,

\[\int{{{\tanh }^{3}}xdx}\]

We know replace \[\tanh x\] with \[\tanh x=\dfrac{\sinh x}{\cosh x}\] and rewrite the given integral as,

\[\Rightarrow \int{\dfrac{{{\sinh }^{3}}x}{{{\cosh }^{3}}x}dx}\]

Now, we substitute an equation for \[{{\sinh }^{2}}x\] . From the general equations of hyperbolic functions, we can write an equation that follows,

\[{{\sinh }^{2}}x={{\cosh }^{2}}x-1\]

Now, replacing this equation in the above integral, we write,

\[\begin{align}

& \Rightarrow \int{\dfrac{\sinh x\cdot {{\sinh }^{2}}x}{{{\cosh }^{3}}x}dx} \\

& =\int{\dfrac{\sinh x\cdot \left( {{\cosh }^{2}}x-1 \right)}{{{\cosh }^{3}}x}dx} \\

\end{align}\]

Now, we assume \[\cosh x=t\] , differentiating both sides we get,

\[\sinh xdx=dt\]

Now, replacing \[\sinh xdx\] with \[dt\] in the above integral, we get,

\[\Rightarrow \int{\dfrac{\left( {{t}^{2}}-1 \right)}{{{t}^{3}}}dt}\]

Now, dividing the above integral by \[{{t}^{3}}\] we get,

\[\Rightarrow \int{\dfrac{1}{t}-\dfrac{1}{{{t}^{3}}}dt}\]

Now, applying the standard formulae for integrals, we can perform the integration as,

\[\begin{align}

& \int{\dfrac{1}{t}dt=\ln \left| t \right|}+c \\

& \int{\dfrac{1}{{{t}^{3}}}dt}=-\dfrac{1}{2{{t}^{2}}}+k \\

\end{align}\]

Now, applying these standard integrals in our problem we get,

\[\Rightarrow \int{\dfrac{1}{t}-\dfrac{1}{{{t}^{3}}}dt}=\ln \left| t \right|+\dfrac{1}{2{{t}^{2}}}+k\]

Now finally replacing the value of \[t\] with \[\cosh x\] we get,

\[\Rightarrow \ln \left| \cosh x \right|+\dfrac{1}{2{{\cosh }^{2}}x}+k\]

Now, replacing \[\operatorname{sech}x=\dfrac{1}{\cosh x}\] we get,

\[\Rightarrow \ln \left| \cosh x \right|+\dfrac{{{\operatorname{sech}}^{2}}x}{2}+k\]

**Note:**For these types of problems, we need to remember about the general formulae and basic equations for hyperbolic functions. Keeping these things in mind, solving the problem becomes very easy and smooth. We should also remember that the solving of these hyperbolic integrals are very similar to that of a normal general trigonometric integral, just there are some differences in the formulae and equations.

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