`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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