Question

$\int\left(\frac{1}{1+e^{t}}\right)dt$ is equal to

• A. $t+\log(1+e^{t})+C$
• B. $t-\log(1-e^{t})+C$
• C. $2t-\log(1+e^{t})+C$
• D. $t-e^{t}\log(1+e^{t})+C$
• E. $t-\log(1+e^{t})+C$

Hint:

Math Tip: Another reliable method for this specific integral is to multiply the numerator and denominator by $e^{-t}$. This transforms the integral to $\int \frac{e^{-t}}{e^{-t} + 1} dt$, which is directly in the form $-\int \frac{f'(t)}{f(t)} dt$, yielding $-\log(e^{-t} + 1)$. This mathematically simplifies to the exact same result!

Answer 1

Answer: (E)

Concept:
Calculus - Integration by Algebraic Manipulation.
Step 1: Manipulate the numerator.
To make the integral solvable, add and subtract $e^t$ in the numerator. This does not change the value of the expression: $$ \int \frac{1 + e^t - e^t}{1 + e^t} dt $$
Step 2: Split the fraction.
Separate the fraction into two parts based on the newly added terms: $$ \int \left( \frac{1 + e^t}{1 + e^t} - \frac{e^t}{1 + e^t} \right) dt $$ $$ \int \left( 1 - \frac{e^t}{1 + e^t} \right) dt $$
Step 3: Integrate the first term.
The integral of a constant is straightforward: $$ \int 1 \,dt = t $$
Step 4: Integrate the second term using substitution.
For the integral $\int \frac{e^t}{1 + e^t} dt$:

• Notice that the numerator is exactly the derivative of the denominator.
• Let $f(t) = 1 + e^t$, then $f'(t) = e^t$.
• Use the standard logarithmic integration rule: $\int \frac{f'(t)}{f(t)} dt = \log|f(t)|$.

$$ \int \frac{e^t}{1 + e^t} dt = \log(1 + e^t) $$
Step 5: Combine the terms.
Combine the results of Step 3 and Step 4, and add the integration constant: $$ t - \log(1 + e^t) + C $$