`int sqrt(5x^2-1)dx` 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 sqrt(5x^2-1) dx` , it resembles one of the formula from integration table.  We may apply the integral formula for function with roots as:

`int sqrt(u^2+-a^2)du=1/2usqrt(u^2+-a^2)+-1/2a^2ln|u+sqrt(u^2+-a^2)| +C` .

Take...

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 sqrt(5x^2-1) dx` , it resembles one of the formula from integration table.  We may apply the integral formula for function with roots as:

`int sqrt(u^2+-a^2)du=1/2usqrt(u^2+-a^2)+-1/2a^2ln|u+sqrt(u^2+-a^2)| +C` .

Take note the sign inside the root is "`(-)` " then we follow the formula as:

`int sqrt(u^2-a^2)du=1/2usqrt(u^2-a^2)-1/2a^2ln|u+sqrt(u^2-a^2)| +C`

By comparing "`u^2-a^2` " with "`5x^2-1` " , we determine the corresponding values as:

`u^2=5x^2` or `(sqrt(5)x)^2` then `u =sqrt(5)x`

`a^2 =1` or `1^2` then `a=1`

For the derivative of `u` , we get `du = sqrt(5) dx` or `(du)/sqrt(5) =dx` .

Plug-in on the values `u^2=5x^2` and `(du)/sqrt(5) =dx` on the integral problem, we get: 

`int sqrt(5x^2-1) dx=int sqrt(u^2-1) *(du)/sqrt(5)`

Apply the basic properties of integration: `int c*f(x) dx= c int f(x) dx` .

`int sqrt(u^2-1) *(du)/sqrt(5) =1/sqrt(5)int sqrt(u^2-1) du`

Apply aforementioned integral formula for function with roots where `a^2 =1` , we get:

`1/sqrt(5)int sqrt(u^2-1) du=1/sqrt(5)*[1/2usqrt(u^2-1)-1/2*1*ln|u+sqrt(u^2-1)|] +C`

                           `=1/sqrt(5)*[1/2usqrt(u^2-1)-1/2ln|u+sqrt(u^2-1)|] +C`

                           `=1/(2sqrt(5))usqrt(u^2-1)-1/(2sqrt(5))ln|u+sqrt(u^2-1)|]+C`

                           `=(usqrt(u^2-1))/(2sqrt(5))- (ln|u+sqrt(u^2-1)|)/(2sqrt(5)) +C`

Plug-in `u^2=5x^2` and `u =sqrt(5)x` on  `(usqrt(u^2-1))/(2sqrt(5))- (ln|u+sqrt(u^2-1)|)/(2sqrt(5)) +C` , we get the indefinite integral as:

`int sqrt(5x^2-1) dx = (sqrt(5)xsqrt(5x^2-1))/(2sqrt(5))- (ln|sqrt(5)x+sqrt(5x^2-1)|)/(2sqrt(5)) +C`

                          `= (xsqrt(5x^2-1))/2- (ln|sqrt(5)x+sqrt(5x^2-1)|)/(2sqrt(5)) +C`