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`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`


`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` .


`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)`


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