Question

solve integral of cosh X/1+e2x​ dx.

Answer 1

$\textbf{To find:}$

$\mathrm{\displaystyle\int\,\dfrac{coshx}{1+e^{2x}}\,dx}$

$\textbf{Solution:}$

$\text{We know that,}$

$\mathrm{coshx=\dfrac{e^x+e^{-x}}{2}}$

$\mathrm{coshx=\dfrac{e^x+\frac{1}{e^x}}{2}}$

$\mathrm{coshx=\dfrac{e^{2x}+1}{2e^x}}$

$\implies\boxed{\mathrm{coshx=\dfrac{1+e^{2x}}{2e^x}}}$

$\text{Now}$

$\mathrm{\displaystyle\int\,\dfrac{coshx}{1+e^{2x}}\,dx}$

$\mathrm{=\displaystyle\int\,\dfrac{coshx}{2\,coshx\,e^{2x}}\,dx}$

$\mathrm{=\displaystyle\int\,\dfrac{1}{2\,e^{2x}}\,dx}$

$\mathrm{=\displaystyle\dfrac{1}{2}\int\,e^{-2x}\,dx}$

$\mathrm{=\dfrac{1}{2}(\dfrac{e^{-2x}}{-2})+C}$

$\mathrm{=\dfrac{e^{-2x}}{-4}+C}$

$\therefore\mathrm{\displaystyle\int\,\dfrac{coshx}{1+e^{2x}}\,dx=\dfrac{e^{-2x}}{-4}+C}$

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