How do you evaluate #int 1/(9+x^2)# from #[sqrt3,3]#?

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Answer 1 · 2017-01-11T16:52:12

#pi/36.#

Let #I=int_sqrt3^3 1/(x^2+9)dx.#

#intf(x)=F(x)+C rArr int_a^bf(x)dx=[F(x)]_a^b=F(b)-F(a)#.

Since, #int1/(x^2+a^2)dx=1/aarc tan(x/a)+c#, we have,

#I=1/3[arc tan(x/3)]_sqrt3^3#

#1/3[arc tan(3/3)-arc tan(sqrt3/3)]#

#=1/3[pi/4-pi/6]#

#:. I=pi/36#.

Enjoy Maths.!