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.!

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