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Evaluate the following $$\int \frac {(\sinh^3 x+1)\cosh x}{\sinh^2 x+1}dx$$

Keep in mind I only learnt the following identity for hyperbolic functions so far $\sinh^2x+1=\cosh^2 x$ and I only know how to differentiate cosh and sinh.

Here is my attempt at a solution

$$\int \frac {(\sinh^3 x+1)\cosh x}{\sinh^2 x+1}dx=\int \frac {(\sinh^3 x+1)\cosh x}{\cosh^2 x}dx=\int \frac {(\sinh^3 x+1)}{\cosh x}dx=\int \frac {(\sinh x+1)(\sinh^2 x - \sinh x +1)}{\cosh x}dx=\int \frac {(\sinh x+1)(\cosh^2 x - \sinh x)}{\cosh x}dx=\int \frac {\cosh^2 x \sinh x-\sinh^2 x+\cosh^2 x - \sinh x}{\cosh x}dx=\int\cosh x \sinh x-\frac{\sinh^2 x}{\cosh x}+\cosh x-\frac{\sinh x}{\cosh x}=0.5\sinh^2 x+\sinh x-\ln|\cosh x|-\int\frac{\sinh^2 x}{\cosh x}dx$$

I find myself unable to integrate $\int\frac{\sinh^2 x}{\cosh x}dx$, Hints only thanks in advance!

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Hint: $\frac{\mathrm{d}}{\mathrm{d}x}\sinh(x)=\cosh(x)$

A change of variables, $u=\sinh(x)$, yields $$ \int\frac{u^3+1}{u^2+1}\mathrm{d}u $$

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  • $\begingroup$ Oh goodness, how could I have been so blind... THANKS! $\endgroup$ – Yellow Skies Nov 11 '12 at 13:40
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Hint: Use the identity $\sinh^2x=\cosh^2x-1$

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  • $\begingroup$ I am trying that on the final equation, which leaves me with $\frac{\cosh^2 x-1}{\cosh x}=\cosh x-\frac{1}{\cosh x}$ I cannot solve $\int \frac{1}{cosh x}$ $\endgroup$ – Yellow Skies Nov 11 '12 at 13:29
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To solve integral of $1/\cosh x$, multiply by $\cosh x /\cosh x$, then convert the denominator to $1+\sinh^2(x)$. It becomes an arctangent. See http://www.integraltec.com/math/math.php?f=sechx.html

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