How do you find the antiderivative of #(e^(2x)+e^x)/(e^(2x)+1)#?

1 Answer
Jul 31, 2016

Answer:

#= 1/2 ln (e^(2x)+1) + arctan e^x + C#

Explanation:

#int \ (e^(2x)+e^x)/(e^(2x)+1) \ dx#

#= int \ (e^(2x))/(e^(2x)+1) + (e^(x))/(e^(2x)+1) \ dx#

spotting the pattern
#= int \ d/dx (1/2 ln (e^(2x)+1)) + (e^(x))/(e^(2x)+1) \ dx#

#= 1/2 ln (e^(2x)+1) + int \ (e^(x))/(e^(2x)+1) \ dx qquad triangle#

for the remaining part we sub #p = e^x, dp = e^x dx = p \ dx#

so #int \ (e^(x))/(e^(2x)+1) \ dx to int (p)/(p^2 + 1) \ 1/p \ dp#

#= int (1)/(p^2 + 1) \ dp = arctan p# where #p = e^x#

so #triangle# becomes

#= 1/2 ln (e^(2x)+1) + arctan e^x + C#