# 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...

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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 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

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