Finding interesting ways to solve $\int \frac{2\sinh(x)+1}{2\cosh(3x)} {\rm d}x$. $$\int \frac{2\sinh(x)+1}{2\cosh(3x)} {\rm d}x
\\ = \int \frac{e^x + 1 - e^{-x}}{e^{3x} + e^{-3x}} {\rm d}x
\\ = \int \frac{e^{4x} + e^{3x} - e^{2x}}{e^{6x} + 1} {\rm d}x
\\ = \int \frac{u^3 + u^2 -u}{u^6+1} {\rm d}u
$$
I do know how to solve the latter one but with really long boring procedures, is there a better way to do this? Splitting it into
$$\int \left( \frac{u^3}{u^6+1} + \frac{u^2}{u^6+1} - \frac{u}{u^6+1} \right) {\rm d}x$$
and using the standard methods can eventually solve it, but it's kinda boring. Like by making substitutions to turn the denominator into cubic/quadratic functions, applying partial fractions, tangent substitutions etc. Can you spot simpler (i.e cooler) ways to evaluate this? Optimally without partial fractions.
 A: $$\int \frac{e^x + 1 - e^{-x}}{e^{3x} + e^{-3x}} {\rm d}x$$
Splitting into two parts: 
$$ \int \frac{e^x - e^{-x}}{e^{3x} + e^{-3x}} {\rm d}x + \int \frac{1}{e^{3x} + e^{-3x}} {\rm d}x $$
Looking at the first part:
$$ \int \frac{e^x - e^{-x}}{e^{3x} + e^{-3x}} {\rm d}x = \int \frac{e^x - e^{-x}}{(e^{x} + e^{-x})^3 - 3 \left (e^x + e^{-x}\right)} {\rm d}x$$
Here let $e^x + \frac{1}{e^x} = t$. 
$$\implies e^x - e^{-x} dx = dt$$
Therefore the first integral now is: 
$$\int \frac{1}{t^3 -3t} {\rm d}t$$
A little bit of partial fractions will be involved but it will only be a single step from completion as compared to the ones you're using right now. 
The second integral, i.e. 
$$\int \frac{1}{e^{3x} + e^{-3x}} {\rm d}x  $$
$$=\int \frac{e^{3x}}{e^{6x} + 1}  {\rm d}x $$
Here simply let $e^{3x} = t$ 
$$\implies 3 e^{3x} dx = dt$$
Therefore, 
$$\int \frac{e^{3x}}{e^{6x} + 1}  {\rm d}x$$
$$ = \int \frac{1}{3(1+t^2)}  {\rm d}t$$
Hope you found it cool.
Cheers!
A: $$
\int{\frac{2\sinh (x)+1}{2\cosh (3x)}dx}=\int{\frac{\sinh (x)}{\cosh (3x)}dx}+\frac{1}{2}\int{\frac{1}{\cosh (3x)}dx}
$$
For the second integral use: ${{\cosh }^{2}}\left( 3x \right)-{{\sinh }^{2}}\left( 3x \right)=1$  to get:
$$\int{\frac{{{\cosh }^{2}}\left( 3x \right)-{{\sinh }^{2}}\left( 3x \right)}{\cosh (3x)}dx}=\int{\cosh \left( 3x \right)-\tanh \left( 3x \right)\operatorname{sech}\left( 3x \right)dx}=\frac{1}{3}\sinh \left( 3x \right)+\frac{1}{3}\operatorname{sech}\left( 3x \right)$$
For the first integral use : $\cosh \left( 3x \right)=4{{\cosh }^{3}}\left( x \right)-3\cosh \left( x \right)$ to get:
$$\int{\frac{\sinh (x)}{4{{\cosh }^{3}}\left( x \right)-3\cosh \left( x \right)}dx}=\int{\frac{d\left( \cosh \left( x \right) \right)}{4{{\cosh }^{3}}\left( x \right)-3\cosh \left( x \right)}}=\int{\frac{du}{4{{u}^{3}}-3u}}$$
A: Let's factorise the denominator, so your integral is $$\int\left(\frac{u^3+u^2-u}{(u^2+1)(u^4-u^2+1)}\right)du=\int\left(\frac{-2u-1}{3(u^2+1)}+\frac{\frac12(4u^3-2u)+u^2+1}{3(u^4-u^2+1)}\right)du\\=-\frac13\ln(u^2+1)-\frac13\arctan u+\frac16\ln(u^4-u^2+1)+\frac13\int\frac{u^2+1}{u^4-u^2+1}du.$$The last integrand is easy to rewrite in partial fractions using $y=u^2$ (don't integrate by substitution!), viz. $$\int\frac{u^2+1}{u^4-u^2+1}du=\frac12\sum_\pm\int\frac{du}{u^2\pm\sqrt{3}u+1},$$which I'm sure you know how to evaluate. But since there are so many arctangents involved, bear in mind trigonometric identities might lead to a different-looking but identical result (to within an integration constant) from other methods, including the "method" of asking software for the answer.
Edit: an alternative approach is to separately evaluate $\int\frac{\sinh x dx}{\cosh 3x}$ and $\frac12\int\operatorname{sech}3xdx$, the former with $c:=\cosh x\implies\cosh 3x=c(4c^2-3)$ and the latter with this.
A: $$
\int \frac{2\sinh(x)+1}{2\cosh(3x)} {\rm d}x =\int \frac{e^{4x} + e^{3x} - e^{2x}}{e^{6x} + 1} {\rm d}x
\\= \int \frac{e^{4x} - e^{2x} + 1}{(e^{2x} + 1)(e^{4x} - e^{2x} + 1)} -
\frac{1}{e^{6x}+1} + \frac{e^{3x}}{e^{6x}+1} {\rm d}x
\\= \int \frac{e^{-2x}}{1+e^{-2x}} {\rm d}x \ - \int\frac{e^{-6x}}{1+e^{-6x}}{\rm d}x \ + \frac13\int \frac{3e^{3x}}{(e^{3x})^2 +1} 
\\= -\frac12 \ln|1+e^{-2x}| + \frac16 \ln|1+e^{-6x}| + \frac13 \arctan(e^{3x}) + C
\\= \boxed{ \frac16 \ln \left( \frac{4\sinh^2(x)+1}{4\cosh^2(x)} \right)+ \frac13 \arctan(e^{3x}) + C}
$$
