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Integrate
1//(1+x^(2)) for limit
[0,1].

Integrate $\frac{1}{1+x^{2}}$ for limit $[0,1]$ .

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Find limits of polynomials and rational functions

Full solution

Q. Integrate

\frac{1}{1+x^{2}}

for limit

[0,1]

.

Find Antiderivative: We need to integrate the function $\frac{1}{1+x^2}$ within the limits from $0$ to $1$ . The antiderivative of $\frac{1}{1+x^2}$ is $\text{arctan}(x)$ , since the derivative of $\text{arctan}(x)$ is $\frac{1}{1+x^2}$ .
Apply Definite Integral: Now we will apply the definite integral with the limits from $0$ to $1$ to the antiderivative we found. $\newline$ $\int_{0}^{1} \frac{1}{1+x^2} dx = [\arctan(x)]_{0}^{1}$
Evaluate at Limits: We will evaluate the antiderivative at the upper limit and then subtract the evaluation at the lower limit. $\newline$ $\arctan(1) - \arctan(0)$
Calculate Final Result: We know that $\text{arctan}(1)$ is $\frac{\pi}{4}$ because $\tan\left(\frac{\pi}{4}\right) = 1$ , and $\text{arctan}(0)$ is $0$ because $\tan(0) = 0$ . So, $\text{arctan}(1) - \text{arctan}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}$
Calculate Final Result: We know that $\arctan(1)$ is $\frac{\pi}{4}$ because $\tan\left(\frac{\pi}{4}\right) = 1$ , and $\arctan(0)$ is $0$ because $\tan(0) = 0$ . So, $\arctan(1) - \arctan(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}$ Therefore, the integral of $\frac{1}{1+x^2}$ from $0$ to $1$ is $\frac{\pi}{4}$ .

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