Question

$\int \frac{dx}{1+e^x}$

Answer 1

$x - \ln(1+e^x) + C$

Answer 2

To solve the integral $\int \frac{dx}{1+e^x}$, we can use a substitution method or algebraic manipulation.

Method 1: Algebraic Manipulation and Substitution

1. Multiply numerator and denominator by $e^{-x}$: $I = \int \frac{dx}{1+e^x} = \int \frac{e^{-x} dx}{e^{-x}(1+e^x)} = \int \frac{e^{-x} dx}{e^{-x}+1}$

2. Perform a substitution: Let $u = e^{-x}+1$. Differentiate $u$ with respect to $x$: $du = -e^{-x} dx$ So, $e^{-x} dx = -du$.

3. Substitute into the integral: $I = \int \frac{-du}{u} = -\int \frac{1}{u} du$

4. Integrate with respect to $u$: $I = -\ln|u| + C$

5. Substitute back $u = e^{-x}+1$: $I = -\ln|e^{-x}+1| + C$ Since $e^{-x} > 0$ for all real $x$, $e^{-x}+1$ is always positive, so we can remove the absolute value signs: $I = -\ln(e^{-x}+1) + C$

6. Simplify the expression: We can rewrite $e^{-x}+1$ as $\frac{1}{e^x}+1 = \frac{1+e^x}{e^x}$. $I = -\ln\left(\frac{1+e^x}{e^x}\right) + C$ Using logarithm properties, $\ln\left(\frac{a}{b}\right) = \ln a - \ln b$: $I = -(\ln(1+e^x) - \ln(e^x)) + C$ $I = -\ln(1+e^x) + \ln(e^x) + C$ Since $\ln(e^x) = x$: $I = x - \ln(1+e^x) + C$

Method 2: Substitution and Partial Fractions

1. Perform a substitution: Let $t = e^x$. Differentiate $t$ with respect to $x$: $dt = e^x dx$ So, $dx = \frac{dt}{e^x} = \frac{dt}{t}$.

2. Substitute into the integral: $I = \int \frac{1}{1+t} \cdot \frac{dt}{t} = \int \frac{1}{t(1+t)} dt$

3. Decompose the integrand using partial fractions: We want to find $A$ and $B$ such that: $\frac{1}{t(1+t)} = \frac{A}{t} + \frac{B}{1+t}$ Multiply both sides by $t(1+t)$: $1 = A(1+t) + Bt$ Set $t=0$: $1 = A(1+0) + B(0) \implies A=1$. Set $t=-1$: $1 = A(1-1) + B(-1) \implies 1 = -B \implies B=-1$. So, the partial fraction decomposition is: $\frac{1}{t(1+t)} = \frac{1}{t} - \frac{1}{1+t}$

4. Substitute back into the integral and integrate: $I = \int \left(\frac{1}{t} - \frac{1}{1+t}\right) dt$ $I = \int \frac{1}{t} dt - \int \frac{1}{1+t} dt$ $I = \ln|t| - \ln|1+t| + C$

5. Substitute back $t = e^x$: $I = \ln|e^x| - \ln|1+e^x| + C$ Since $e^x > 0$ and $1+e^x > 0$, we can remove the absolute value signs: $I = \ln(e^x) - \ln(1+e^x) + C$ Since $\ln(e^x) = x$: $I = x - \ln(1+e^x) + C$

Both methods yield the same result.