# int 1/(x^2sqrt(x^2-4)) dx Use integration tables to find the indefinite integral. Recall that indefinite integral follows the formula: int f(x) dx = F(x) +C

where: f(x) as the integrand

F(x) as the anti-derivative function

C  as the arbitrary constant known as constant of integration

For the given problem int 1/(x^2sqrt(x^2-4)) dx...

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Recall that indefinite integral follows the formula: int f(x) dx = F(x) +C

where: f(x) as the integrand

F(x) as the anti-derivative function

C  as the arbitrary constant known as constant of integration

For the given problem int 1/(x^2sqrt(x^2-4)) dx , it resembles one of the formula from integration table.  We may apply the integral formula for rational function with roots as:

int 1/(u^2sqrt(u^2-a^2))du = 1/(a^2*u) sqrt(u^2-a^2)+C .

By comparing "u^2-a^2 " with "x^2-4 " , we determine the corresponding values as:

u^2=x^2  then u =x

a^2 =4

Plug-in the values on the aforementioned integral formula for rational function with roots where  a^2 =4 , we get:

int 1/(x^2sqrt(x^2-4)) dx=1/(4*x) sqrt(x^2-4)+C

 =1/(4x) sqrt(x^2-4)+C

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