Question

Integrate 1/(e^2x+e^x)dx

Answer 1

ln(1 + e^{-x}) - e^{-x} + C

Answer 2

To integrate the given function $\frac{1}{e^{2x} + e^x}$, we follow these steps:

1. Substitution: Let the integral be $I = \int \frac{1}{e^{2x} + e^x} dx$. First, factor out $e^x$ from the denominator: $I = \int \frac{1}{e^x(e^x + 1)} dx$. Now, let $u = e^x$. Then, differentiate $u$ with respect to $x$: $du = e^x dx$. From this, we can express $dx$ in terms of $du$ and $u$: $dx = \frac{du}{e^x} = \frac{du}{u}$.

2. Transform the Integral: Substitute $u$ and $dx$ into the integral: $I = \int \frac{1}{u(u+1)} \cdot \frac{du}{u}$ $I = \int \frac{1}{u^2(u+1)} du$.

3. Partial Fraction Decomposition: The integrand is a rational function, so we use partial fraction decomposition. Let $\frac{1}{u^2(u+1)} = \frac{A}{u} + \frac{B}{u^2} + \frac{C}{u+1}$. To find the constants $A$, $B$, and $C$, multiply both sides by $u^2(u+1)$: $1 = A u(u+1) + B(u+1) + C u^2$.

   • Set $u=0$: $1 = A(0) + B(0+1) + C(0)^2$ $1 = B \implies B=1$.

   • Set $u=-1$: $1 = A(-1)(-1+1) + B(-1+1) + C(-1)^2$ $1 = A(0) + B(0) + C(1)$ $1 = C \implies C=1$.

   • Set $u=1$ (or any other convenient value) and substitute the values of $B$ and $C$: $1 = A(1)(1+1) + B(1+1) + C(1)^2$ $1 = 2A + 2B + C$ $1 = 2A + 2(1) + 1$ $1 = 2A + 2 + 1$ $1 = 2A + 3$ $2A = -2 \implies A=-1$.

   So, the partial fraction decomposition is: $\frac{1}{u^2(u+1)} = -\frac{1}{u} + \frac{1}{u^2} + \frac{1}{u+1}$.

4. Integrate: Now, integrate each term: $I = \int \left( -\frac{1}{u} + \frac{1}{u^2} + \frac{1}{u+1} \right) du$ $I = -\int \frac{1}{u} du + \int u^{-2} du + \int \frac{1}{u+1} du$ $I = -\ln|u| + \frac{u^{-1}}{-1} + \ln|u+1| + C'$ $I = -\ln|u| - \frac{1}{u} + \ln|u+1| + C'$. Combine the logarithmic terms: $I = \ln\left|\frac{u+1}{u}\right| - \frac{1}{u} + C'$.

5. Substitute Back: Substitute $u = e^x$ back into the expression. Since $e^x > 0$ for all real $x$, $|e^x| = e^x$ and $|e^x+1| = e^x+1$. $I = \ln\left(\frac{e^x+1}{e^x}\right) - \frac{1}{e^x} + C'$ $I = \ln\left(1 + \frac{1}{e^x}\right) - e^{-x} + C'$ $I = \ln(1 + e^{-x}) - e^{-x} + C'$.