then you can use partial fraction i.e. create (t)- (t-1) in the numeratorHallsofIvy said:You have done the substitution wrong.
If [itex]t= 1+ e^x[/itex] then [itex]dt= e^xdx= (t- 1)dx[/itex] so that [itex]\frac{1}{t- 1}= dx[/itex].
[tex]\int \frac{1}{1+ e^x}dx= \int \frac{1}{t} \frac{dt}{t- 1}= \int \frac{1}{t(t- 1)} dt[/tex]
NOT [itex]\int \frac{1}{t^2- 1} dt[/itex]
The general approach to solving this integral is to use the substitution method. We substitute u = e^x and then use the fact that du/dx = e^x to rewrite the integral in terms of u. This will allow us to use the formula for the integral of 1/u, which will help us solve the problem.
To integrate 1/(1+e^x), we first use the substitution method to rewrite the integral in terms of u = e^x. This will give us the integral of 1/u, which can be solved using the formula for integrating 1/u. After solving for the integral in terms of u, we can then substitute back in e^x for u to get the final solution.
Yes, there is a shortcut or trick to solving this integral. We can use the fact that 1/(1+e^x) can be rewritten as e^x/(1+e^x). This will allow us to use the formula for the integral of e^x/(1+e^x), which is ln(1+e^x). This shortcut can save time and effort when solving the integral.
There are no special cases or exceptions when solving this integral. The substitution method and the shortcut mentioned above can be used for all cases of the integral 1/(1+e^x).
We can check our solution by taking the derivative of the final answer. If the derivative matches the original integrand, then our solution is correct. In this case, we can also check our solution by plugging in different values of x and seeing if the result matches the original integrand of 1/(1+e^x).