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

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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...

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