MHB Integration Problem - Can Someone Help?

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The discussion centers on solving an integration problem involving hyperbolic functions. The original integral is transformed using the identity f'(x)/f(x), leading to a breakdown into simpler components. Participants are collaborating to simplify the integral further, with suggestions to separate terms and apply known integration techniques. The conversation emphasizes the need for clarity in each step of the integration process. The thread remains focused on finding a complete solution to the integral presented.
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Can someone help with this problem, i tried to solve it
using f'(x)/f(x) but couldn't figure it out.View attachment 5909
 

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Hi mahmoud shaaban and welcome to MHB! :D

$$\begin{align*}\int\dfrac{6\sinh(x)\cosh^2(x)+\sinh(x)}{\cosh^3(x)+\cosh(x)}\,dx&=\int\tanh(x)\left(\dfrac{5\cosh^2(x)}{\cosh^2(x)+1}+1\right)\,dx \\
&=5\int\dfrac{\cosh(x)\sinh(x)}{\cosh^2(x)+1}\,dx+\int\tanh(x)\,dx\end{align*}$$

Can you continue?
 
mahmoud shaaban said:
\displaystyle \int \frac{6\sinh x\cosh^2x + \sinh x}{\cosh^3x + \cosh x}\,dx
\text{We have: }\;\int \left(\frac{3\sinh x\cosh^2\!x + \sinh x}{\cosh^3\!x + \cosh x} + \frac{3\sinh x\cosh^2\!x}{\cosh^3\!x + \cosh x}\right)\,dx

. . . =\;\int\left(\frac{3\sinh x\cosh^2\!x + \sinh x}{\cosh^3\!x + \cosh x} +\frac{3\sinh x \cosh x}{\cosh^2\!x + 1} \right)\,dx

. . . =\;\int \frac{3\sinh x\cosh^2\!x + \sinh x}{\cosh^3\!x + \cosh x}\,dx \;+\; \frac{3}{2}\int\frac{2\sinh x\cosh x}{\cosh^2\!x+1}\,dx

Can you finish it now?
 
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