Final answer:
The value of the improper integral ∫(1 to ∞) 1/(x^2+1) dx is π/4. The calculation involves evaluating the function arctan(x), utilizing the Fundamental Theorem of Calculus, and taking the limit as the upper bound approaches infinity.
Step-by-step explanation:
To compute the value of the improper integral ∫(1 to ∞) 1/(x^2+1) dx, we need to evaluate the definite integral as a limit:
\[\int_{1}^{\infty} \frac{1}{x^2+1} dx = \lim_{b \to \infty} \int_{1}^{b} \frac{1}{x^2+1} dx\]
We recognize that the integrand 1/(x^2+1) is the derivative of arctan(x), so we can integrate using a substitution method:
\[\int \frac{1}{x^2+1} dx = \arctan(x) + C\]
Then, applying the Fundamental Theorem of Calculus, we have:
\[\lim_{b \to \infty} (\arctan(b) - \arctan(1))\]
Since \(\arctan(\infty)\) is \(\frac{\pi}{2}\) and \(\arctan(1)\) is \(\frac{\pi}{4}\), we find:
\[\lim_{b \to \infty} (\arctan(b) - \arctan(1)) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}\]
Therefore, the value of the improper integral is \(\frac{\pi}{4}\).