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

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Answer 1 · 2016-08-01T14:35:22

#pi/2#.

Let #I=int_-1^1 dx/(1+x^2)#.

Then, #I=[arctanx]_-1^1#

#=arctan1-arctan(-1)#

#=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#.