How do you find the definite integral for: #2 / (4+x^2) dx# for the intervals #[0, 2]#?

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Answer 1 · 2016-07-09T21:27:26

#int_0^2 (2)/(4+x^2)dx =2* int_0^2 (1)/(2^2+x^2)dx#

Knowing that

#int 1/(a^2+x^2)dx = 1/a * arctan(x/a)+C#

We see that

#2* int_0^2 (1)/(2^2+x^2)dx = 2*[1/2 arctan(2/2) - 1/2 arctan(0/2)]#

#=arctan(1)-arctan(0)=arctan(1)=pi/4#

For this problem, it is good to know some basic integral formulas such as

#int 1/(a^2+x^2)dx = 1/a * arctan(x/a)+C#