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

1 Answer
Jul 31, 2016

= 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