int sqrt(25x^2+4)/x^4 dx Find the indefinite integral

To solve the indefinite integral, we follow int f(x) dx = F(x) +C
where:
f(x) as the integrand function
F(x) as the antiderivative of f(x)
C as the constant of integration.
For the given integral problem: int sqrt(25x^2+4)/x^4dx , we may apply integration by parts: int u *dv = uv - int v *du .
Let:
u =sqrt(25x^2+4)
Apply Law of Exponent: sqrt(x) = x^(1/2) , we get:  u =(25x^2+4)^(1/2)
To find the derivative of u , we may apply Power rule for derivative: d/(dx) u^n= n* u^(n-1) * d/(dx)(u)
u' = 1/2 *(25x^2+4)^(1/2-1) * d/(dx) (25x^2+4)
u' = 1/2*(25x^2+4)^(-1/2)* (50xdx)
u' = 25x(25x^2+4)^(-1/2)dx
Apply Law of exponent: 1/x^n =x^(-n) .
u' = (25x)/(25x^2+4)^(1/2)dx or(25x)/sqrt(25x^2+4)dx
 Let: v' = 1/x^4 dx
To find the integral of v' , we apply  Law of exponent: 1/x^n =x^(-n) and Power rule for integration:int x^n dx = x^(n+1)/(n+1) +C .
v = int v'
  = int 1/x^4 dx
  = int x^(-4) dx
  = x^(-4+1)/(-4+1)
  = x^(-3)/(-3)
  = - 1/(3x^3)
Apply the formula for integration by parts using the following values: u =sqrt(25x^2+4) , u' =(25x)/sqrt(25x^2+4) dx , v'= 1/x^4 dx and v=- 1/(3x^3) .
int sqrt(25x^2+4)/x^4dx =sqrt(25x^2+4)*-( 1/(3x^3)) - int(25x)/sqrt(25x^2+4)*(- 1/(3x^3))dx
                              =-sqrt(25x^2+4)/(3x^3) - int -25/(3x^2sqrt(25x^2+4))dx
To evaluate the integral part, we may apply the basic integration property: int c*f(x) dx = c int f(x) dx .
int -(25x)/(3x^3sqrt(25x^2+4))dx =-25/3int 1/(x^2sqrt(25x^2+4))dx
The integral resembles one of the formulas from the integration table for rational function with roots. We follow:
int (du)/(u^2sqrt(u^2+a^2)) = -sqrt(u^2+a^2)/(a^2x) +C
For easier comparison, we may apply u-substitution by letting: u^2 = 25x^2 or (5x)^2 then u=5x and du = 5dx or (du)/5 = dx . When we let u^2 =25x^2 , it can be rearrange as x^2=u^2/25 . Applying the values, the integral becomes:
-25/3int 1/(x^2sqrt(25x^2+4))dx =-25/3int 1/((u^2/25)sqrt(u^2+4))*(du)/5
                                         =-25/3int 25/(5u^2sqrt(u^2+4))du
                                        = -125/3int 1/(u^2sqrt(u^2+4))dx
By comparing "u^2sqrt(u^2+a^2) " with "u^2sqrt(u^2+4) ", we determine the corresponding value: a^2=4 . Applying the aforementioned integration formula for rational function with roots, we get:
-125/3int 1/(u^2sqrt(u^2+4))dx =-125/3* [-sqrt(u^2+4)/(4u)] +C
                                      =(125sqrt(u^2+4))/(12u) +C
Plug-in u^2= 25x^2 and u =5x on  (125sqrt(u^2+4))/(12u) +C , we get the indefinite integral:
int -25/(3x^2sqrt(25x^2+4))dx=(125sqrt(25x^2+4))/(12*5x) +C
                                       =(25sqrt(25x^2+4))/(12x) +C .
For the complete indefinite integral, we get:
int sqrt(25x^2+4)/x^4dx =-sqrt(25x^2+4)/(3x^3) - int -25/(3x^2sqrt(25x^2+4))dx
                             =-sqrt(25x^2+4)/(3x^3)-(25sqrt(25x^2+4))/(12x) +C