How to evaluate #int_1^2(e^(2x)/(e^x-1))# ? Can you solve it without substitution and somehow use chain rule?

Question

1016 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-14T13:59:54

#int_1^2 e^(2x)/(e^x-1)dx = e(e-1)+ln(e+1)#

#int_1^2 e^(2x)/(e^x-1)dx = int_1^2 e^x/(e^x-1)e^xdx#

#int_1^2 e^(2x)/(e^x-1)dx = int_1^2 e^x/(e^x-1)d(e^x)#

#int_1^2 e^(2x)/(e^x-1)dx = int_1^2 (e^x-1+1)/(e^x-1)d(e^x)#

#int_1^2 e^(2x)/(e^x-1)dx = int_1^2 (1+1/(e^x-1))d(e^x)#

#int_1^2 e^(2x)/(e^x-1)dx = int_1^2d(e^x) + int_1^2 1/(e^x-1)d(e^x)#

#int_1^2 e^(2x)/(e^x-1)dx = int_1^2d(e^x) + int_1^2 (d(e^x-1))/(e^x-1)#

#int_1^2 e^(2x)/(e^x-1)dx = [e^x+ln(e^x-1)]_1^2#

#int_1^2 e^(2x)/(e^x-1)dx = e^2+ln(e^2-1)-e-ln(e-1)#

#int_1^2 e^(2x)/(e^x-1)dx = e(e-1)+ln((e^2-1)/(e-1))#

#int_1^2 e^(2x)/(e^x-1)dx = e(e-1)+ln(e+1)#