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How to evaluate #int_0^2(e^(2x)/(e^x-1))#? Is there a way of not directly doing substitution?

1 Answer

Jun 13, 2018

#int_0^2 (e^(2x)*dx)/(e^x-1)=oo#

Explanation:

#int_0^2 (e^(2x)*dx)/(e^x-1)#

=#int_0^2 (e^x*e^x*dx)/(e^x-1)#

After using #u=e^x-1# and #du=e^x*dx# transforms,

#int_0^(e^2-1) ((u+1)*du)/u#

=#oo#

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