Calculus of a Single Variable, Chapter 8, 8.4, Section 8.4, Problem 20

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