How do you find the antiderivative of # {(e^x)/ [(e^(2x)) - 1]}#?

1 Answer
Oct 22, 2016

Answer:

#ln(sqrt(e^(2x)-1)/(e^x+1))+C#

Explanation:

#I=inte^x/(e^(2x)-1)dx#

Let #u=e^x# so that #du=e^xdx#:

#I=inte^x/((e^x)^2-1)dx=int1/(u^2-1)du#

Now, we can ley #u=sectheta#. This implies that #du=secthetatanthetad theta#.

Plugging these in:

#I=int(secthetatantheta)/(sec^2theta-1)d theta#

Note that #1+tan^2theta=sec^2theta#, so #sec^2theta-1=tan^2theta#:

#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 #u=sectheta#. This means we have a right triangle where #u# is the hypotenuse, #1# is the side adjacent to #theta#, and #sqrt(u^2-1)# is the side opposite #theta#.

Thus #csctheta=u/sqrt(u^2-1)# and #cottheta=1/sqrt(u^2-1)#.

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 #1# as an exponent through the rule #Blog(A)=log(A^B)#:

#I=ln(abs(sqrt(u^2-1)/(u+1)))+C#

Through #u=e^x#:

#I=ln(abs(sqrt(e^(2x)-1)/(e^x+1)))+C#

Notice that #sqrt(e^(2x)-1)>0# and #e^x+1>0# for all values of #x#, so the absolute value bars can be removed:

#I=ln(sqrt(e^(2x)-1)/(e^x+1))+C#