Question

The value of
integral of dx/(1+e^x)(1+e^-x)​

Answer 1

Answer:

Step-by-step explanation:

Whenever you see an exponential or a square root, it’s usually fruitful to make a substitution that involves the exponential or root, because the thing you’re substituting is contained within its own derivative. So, we’ll make a substitution of  u=1+ex⟹du=ex dx=(u−1) dx :

I=∫du(u−1)u

Then, we can try to simplify using partial fractions, by trying to get the integrand into the following form:

1u(u−1)=Au+Bu−1

1=A(u−1)+Bu=(A+B)u−A⟹A=−1, B=1

Hence, our integral becomes:

I=∫1u−1−1u du=ln|u−1|−ln|u|+C

I=x−ln(1+ex)+C

The absolute value is dropped because  1+ex  is positive for all real  x .

I=∫11+exdx  

=∫exex(1+ex)dx  

Let's assume  ex=u⇒exdx=du  

I=∫1u(u+1)du  

I=∫(1u−1u+1)du  

I=ln|u|−ln|u+1|+C     , C is a constant of integration.

I=ln|uu+1|+C  

I=lnexex+1+C  

I=x−ln(ex+1)+C  

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