Question #68d98
3 Answers
The answer is
Explanation:
Start by letting
Next, use the Method of Partial Fractions to write
Now substitute
This can be checked by differentiation:
Explanation:
Given:
Insert zero into the numerator:
In place of the 0 we write
Now we do some clever grouping:
We combine the first group:
Separate into two fractions:
Please notice that the second fraction becomes 1:
Separate into two integrals:
For the first integral we let
We know these integrals very well:
Reverse the substitution:
Explanation:
We can also do this integral as follows:
int(e^x-1)/(e^x+1)dx=(e^x+1-2)/(e^x+1)dx
Splitting up the fraction as
=int(1-2/(e^x+1))dx
We can integrate the first term easily:
=x-2int1/(e^x+1)dx
Now, multiply the integrand by
=x-2int(e^-x)/(1+e^-x)dx
Let
=x+2int1/udu
=x+2lnabsu+C
=x+2lnabs(1+e^-x)+C
As
x+2ln(1+e^-x)+C
Which can be shown to be equivalent to the other provided answers.