What is the definite integral of $1/(36+x^2)$ with bounds $[0, 6]$ ?

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Answer 1 · 2016-01-10T20:59:30

$pi/24$

There is a standard integral: $int(1/(a^2+x^2))dx = 1/atan^-1(x/a)+c$

So $int_0^6(1/(36+x^2))dx = [1/6tan^-1(x/6)]_0^6$

$=(1/6tan^-1(1))-(1/6tan^-1(0)) = 1/6*pi/4 - 1/6*0 = pi/24$

What is the definite integral of 1/(36+x^2)1/(36+x^2) with bounds [0, 6][0, 6]?