Question

what is the integreation of dx/1+e^-x.​

Answer 1

Answer:

A lot of good answers here, but I have another approach:

∫11+exdx

We will rewrite our numerator as:

1+ex−ex

This rewriting of the numerator is algebraically legal as the adding and subtracting of a value ultimately results in a change of 0 value.

By doing this, our integral can be written as:

∫1+ex−ex1+exdx

∫1+ex1+exdx+∫−ex1+exdx

First integral:

∫1+ex1+exdx

Since the numerator and denominator are exactly the same, our integrand simplifies to 1 and our integral becomes:

∫dx

Integrating using the power rule of integration:

∫xndx=xn+1n+1

Therefore:

∫1+ex1+exdx=x

Second integral:

∫−ex1+exdx

Let u=1+ex , du=exdx

Apply u -substitution:

∫1u(−du)

−∫1u(du)=−ln|u|

Undo the substitution:

∫−ex1+exdx=−ln|1+ex|+C

Combining these values, we get:

∫1+ex1+exdx+∫−ex1+exdx=x−ln|1+ex|+C

Check by differentiating:

ddx[x−ln|1+ex|+C]

=1−11+ex(ex)

=1−ex1+ex

=1+ex1+ex−ex1+ex

=1+ex−ex1+ex

=11+ex

Therefore:

∫1+ex1+exdx+∫−ex1+exdx=x−ln|1+ex|+C