How do you evaluate the integral of $\int \frac{d x}{1 + x^{2}}$ $int dx/(1+x^2)$ from -1 to 1?

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Answer 1 · 2016-08-01T14:35:22

$\frac{π}{2}$ $pi/2$ .

Let $I = \int_{- 1}^{1} \frac{d x}{1 + x^{2}}$ $I=int_-1^1 dx/(1+x^2)$ .

Then, $I = {[arctan x]}_{- 1}^{1}$ $I=[arctanx]_-1^1$

$= arctan 1 - arctan (- 1)$ $=arctan1-arctan(-1)$

$= \frac{π}{4} - (- \frac{π}{4})$ $=pi/4-(-pi/4)$

$:. I=pi/2$ .

In an another way, we observe that the function $f(x)=1/(1+x^2)$ ,

is an even function, so, by,

The Rule $: int_-a^af(x)dx=2int_0^af(x)dx, if f$ is even.

$:. I=2int_0^1dx/(1+x^2)=2[arctanx]_0^1=2[pi/4-0]=pi/2$ .

How do you evaluate the integral of int dx/(1+x^2)∫dx1+x2int dx/(1+x^2) from -1 to 1?