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

3 Answers
Feb 3, 2018

Answer:

#arctan(e^x) + C#

Explanation:

#"write "e^x "dx as "d(e^x)" , then we obtain"#
#int (d(e^x))/(1+(e^x)^2)#
#"with the substitution y = "e^x", we get"#
#int (d(y))/(1+y^2)#
#"which is equal to"#

#arctan(y) + C#

#"Now substitute back "y = e^x :#

#arctan(e^x) + C#

Feb 3, 2018

Answer:

#int e^x/(1+e^(2x)) "d"x=arctane^x +"c"#

Explanation:

We want to find #inte^x/(1+e^(2x))"d"x=int1/(1+(e^x)^2)e^x"d"x#

Now let #u=e^x# and so taking the differential on both sides gives #du=e^xdx#. Now we substitute both of these equations into the integral to get

#int1/(1+u^2)"d"u#

This is a standard integral which evaluates to #arctanu#. Substituting back for #x# we get a final answer:

#arctan e^x +"c"#

Feb 3, 2018

Answer:

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

Explanation:

First, we let #u=1+e^(2x)#. To integrate with respect to #u#, we divide by the derivative of #u#, which is #2e^(2x)#:
#int\ e^x/(1+e^(2x))\ dx=1/2int\ e^x/(e^(2x)*u)\ du=1/2int\ e^x/(e^x*e^x*u)\ du=#

#=1/2int\ 1/(e^x*u)\ du#
To integrate with respect to #u#, we need everything expressed in terms of #u#, so we need to solve for what #e^x# is in terms of #u#:
#u=1+e^(2x)#

#e^(2x)=u-1#

#2x=ln(u-1)#

#x=1/2ln(u-1)#

#x=ln((u-1)^(1/2))=ln(sqrt(u-1))#

#e^x=e^(ln(sqrt(u-1)))=sqrt(u-1)#

Now we can plug this back into the integral:
#=1/2int\ 1/(e^x*u)\ du=1/2int\ 1/(sqrt(u-1)*u)\ du#

Next we will introduce a substitution with #z=sqrt(u-1)#. The derivative is:
#(dz)/(du)=1/(2sqrt(u-1)#
so we divide by it to integrate with respect to #z# (remember that dividing is the same as multiplying by the reciprocal):
#1/2int\ 1/(sqrt(u-1)*u)\ du=1/2int\ 1/(sqrt(u-1)*u)*2sqrt(u-1)\ dz=#

#=2/2int\ 1/u\ dz#

Now, we once again we have the wrong variable, so we need to solve for what #u# is equal to in terms of #z#:
#z=sqrt(u-1)#

#u-1=z^2#

#u=z^2+1#

This gives:
#int\ 1/u\ dz=int\ 1/(1+z^2)\ dz#

This is the common derivative of #tan^-1(z)#, so we get:
#int\ 1/(1+z^2)\ dz=tan^-1(z)+C#

Undoing all the substitutions, we get:
#tan^-1(z)+C=tan^-1(sqrt(u-1))+C=#

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

#=tan^-1(e^x)+C#