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

1 Answer
Jul 9, 2016

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

Explanation:

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