int_(-oo)^(oo)(1)/(1+x^(2))dx

Recognize Derivative Relationship: We recognize that the function 1/(1+x^2) is the derivative of arctan(x). Therefore, we can use the antiderivative of 1/(1+x^2) to evaluate the integral. Antiderivative: ∫(1/(1+x^2))dx = arctan(x) + CEvaluate Antiderivative Limits: We need to evaluate the antiderivative from -∞ to ∞. So we calculate the limit as x approaches ∞ and as x approaches -∞ of arctan(x). lim_(x→∞) arctan(x) = π/2 lim_(x→-∞) arctan(x) = -π/2Find Definite Integral: Now we can find the definite integral by subtracting the value of the antiderivative at the lower limit from the value at the upper limit. ∫(-∞)^(∞)(1/(1+x^2))dx = arctan(∞) - arctan(-∞) = π/2 - (-π/2) = π/2 + π/2 = π

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$\int_{(-\infty)}^{(\infty)}\frac{1}{1+x^{2}}dx$

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Calculus

Evaluate definite integrals using the chain rule

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\int_{(-\infty)}^{(\infty)}\frac{1}{1+x^{2}}dx

Recognize Derivative Relationship: We recognize that the function $\frac{1}{1+x^2}$ is the derivative of $\text{arctan}(x)$ . Therefore, we can use the antiderivative of $\frac{1}{1+x^2}$ to evaluate the integral. $\newline$ Antiderivative: $\int(\frac{1}{1+x^2})dx = \text{arctan}(x) + C$
Evaluate Antiderivative Limits: We need to evaluate the antiderivative from $-\infty$ to $\infty$ . So we calculate the limit as $x$ approaches $\infty$ and as $x$ approaches $-\infty$ of $\text{arctan}(x)$ . $\lim_{x\to\infty} \text{arctan}(x) = \frac{\pi}{2}$ $\lim_{x\to-\infty} \text{arctan}(x) = -\frac{\pi}{2}$
Find Definite Integral: Now we can find the definite integral by subtracting the value of the antiderivative at the lower limit from the value at the upper limit. $\newline$ $\int_{(-\infty)}^{(\infty)}\frac{1}{1+x^2}\,dx = \text{arctan}(\infty) - \text{arctan}(-\infty)$ $\newline$ $= \frac{\pi}{2} - \left(-\frac{\pi}{2}\right)$ $\newline$ $= \frac{\pi}{2} + \frac{\pi}{2}$ $\newline$ $= \pi$