int_(-oo)^(oo)(e^(x))/((1+e^(x)))dx

Substitution and Simplification: Simplify the integral using a substitution. Let u = e^x, then du = e^x dx, or dx = du/u. Substitute into the integral: int_(-oo)^(oo)(e^(x))/((1+e^(x)))dx = int_(0)^(oo)1/(1+u) * (du/u)Evaluate Integral with New Variable: Evaluate the integral with the new variable. The limits change as x approaches -∞, u approaches 0, and as x approaches ∞, u approaches ∞. So, int_(0)^(oo)1/(1+u) * (du/u) = int_(0)^(oo)1/(u(1+u))duDecompose Fraction: Decompose the fraction. 1/(u(1+u)) can be written as A/u + B/(1+u). Solving for A and B: 1 = A(1+u) + Bu Setting u=0, A = 1. Setting u=-1, B = -1 (which is incorrect, should be B = 0).

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Q. Evaluate the integral :

\int_{-\infty}^{\infty} \frac{e^{x}}{\left(1+e^{x}\right)} d x

Substitution and Simplification: Simplify the integral using a substitution. $\newline$ Let $u = e^x$ , then $du = e^x dx$ , or $dx = \frac{du}{u}$ . $\newline$ Substitute into the integral: $\newline$ $\int_{(-\infty)}^{(\infty)}\frac{e^{x}}{(1+e^{x})}dx = \int_{0}^{\infty}\frac{1}{(1+u)} \cdot \frac{du}{u}$
Evaluate Integral with New Variable: Evaluate the integral with the new variable. $\newline$ The limits change as $x$ approaches $-\infty$ , $u$ approaches $0$ , and as $x$ approaches $\infty$ , $u$ approaches $\infty$ . $\newline$ So, $\int_{0}^{\infty}\frac{1}{1+u} \cdot \left(\frac{du}{u}\right) = \int_{0}^{\infty}\frac{1}{u(1+u)}du$
Decompose Fraction: Decompose the fraction. $\newline$ $\frac{1}{u(1+u)}$ can be written as $\frac{A}{u} + \frac{B}{1+u}$ . $\newline$ Solving for $A$ and $B$ : $\newline$ $1 = A(1+u) + Bu$ $\newline$ Setting $u=0$ , $A = 1$ . $\newline$ Setting $u=-1$ , $B = -1$ (which is incorrect, should be $B = 0$ ).