How do you find the antiderivative of # {(e^x)/ [(e^(2x)) - 1]}#?
1 Answer
Explanation:
#I=inte^x/(e^(2x)-1)dx#
Let
#I=inte^x/((e^x)^2-1)dx=int1/(u^2-1)du#
Now, we can ley
Plugging these in:
#I=int(secthetatantheta)/(sec^2theta-1)d theta#
Note that
#I=int(secthetatantheta)/tan^2thetad theta#
#I=intsectheta/tanthetad theta#
#I=int1/costheta(costheta/sintheta)d theta#
#I=intcsctheta#
This is a fairly common integral:
#I=-ln(abs(csctheta+cottheta))#
We need to rewrite this using
Thus
So:
#I=-ln(abs(u/sqrt(u^2-1)+1/sqrt(u^2-1)))+C#
#I=-ln(abs((u+1)/sqrt(u^2-1)))+C#
Bringing in the negative
#I=ln(abs(sqrt(u^2-1)/(u+1)))+C#
Through
#I=ln(abs(sqrt(e^(2x)-1)/(e^x+1)))+C#
Notice that
#I=ln(sqrt(e^(2x)-1)/(e^x+1))+C#