$\int \frac{1}{\left(e^{x}+1\right)^{2}} d x$

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∻Solution by Steps∻ ‖ step 1 ⋮ We start with the integral $\int \frac{1}{(e^x + 1)^2} \, dx$. ‖ ‖ step 2 ⋮ To solve this integral, we use the substitution $u = e^x + 1$. Then, $du = e^x \, dx$ and $dx = \frac{du}{e^x}$. ‖ ‖ step 3 ⋮ Since $e^x = u - 1$, we have $dx = \frac{du}{u - 1}$. Substituting these into the integral, we get: $$\int \frac{1}{u^2} \cdot \frac{du}{u - 1} = \int \frac{du}{u^2 (u - 1)}.$$ ‖ ‖ step 4 ⋮ We can decompose the integrand using partial fractions: $$\frac{1}{u^2 (u - 1)} = \frac{A}{u} + \frac{B}{u^2} + \frac{C}{u - 1}.$$ ‖ ‖ step 5 ⋮ Solving for $A$, $B$, and $C$, we get: $$1 = A u (u - 1) + B (u - 1) + C u^2.$$ ‖ ‖ step 6 ⋮ By comparing coefficients, we find: $$A = 1, \quad B = -1, \quad C = 0.$$ ‖ ‖ step 7 ⋮ Substituting back, we have: $$\int \left( \frac{1}{u} - \frac{1}{u^2} \right) \, du.$$ ‖ ‖ step 8 ⋮ Integrating term by term, we get: $$\int \frac{1}{u} \, du - \int \frac{1}{u^2} \, du = \ln|u| + \frac{1}{u} + C.$$ ‖ ‖ step 9 ⋮ Substituting back $u = e^x + 1$, we get: $$\ln|e^x + 1| + \frac{1}{e^x + 1} + C.$$ ‖ ∻Answer∻ ⚹ $\ln|e^x + 1| + \frac{1}{e^x + 1} + C$ ⚹ ∻Key Concept∻ ⚹Substitution and Partial Fractions⚹ ∻Explanation∻ ⚹We used substitution to simplify the integral and then applied partial fractions to decompose the integrand, making it easier to integrate term by term.⚹◊What is the antiderivative of $\frac{1}{(e^x+1)^2}$ with respect to $x$?⍭ Generate me a similar question◊

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