`int 1/(x^2sqrt(x^2-4)) dx` Use integration tables to find the indefinite integral.

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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`...

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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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