How to Find the Indefinite Integral for (4x^2+2√x+1)/(2x√x)?

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SUMMARY

The indefinite integral of the function (4x^2 + 2√x + 1)/(2x√x) is calculated as ∫(4x^2 + 2√x + 1)/(2x√x) dx. The solution simplifies to 4/3√(x^3) + ln(x) - 1/√x + C. The process involves breaking down the integral into manageable parts, specifically integrating terms like 2∫x^(1/2) dx, ∫x^(-1) dx, and 1/2∫x^(-3/2) dx. Verification of the solution can be done by taking the derivative of the result.

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KMcFadden
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Homework Statement


∫▒〖(4x^2+2√x+1)/(2x√x) dx〗


Homework Equations





The Attempt at a Solution


∫▒〖(4x^2+2√x+1)/(2x√x) dx〗
∫▒((4x^2)/(2x^(3/2) )+(2√x)/(2x^(3/2) )+1/(2x^(3/2) ))dx
2∫▒x^(1/2) dx+∫▒〖x^(-1) dx〗+1/2 ∫▒x^(-3/2) dx
2×2/3 x^(3/2)+ln⁡x+1/2×(-2) x^(-1/2)+C
4/3 x^(3/2)+ln⁡x-x^(-1/2)+C
4/3 √(x^3 )+ln⁡x-1/√x+C
 
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KMcFadden said:

Homework Statement


∫▒〖(4x^2+2√x+1)/(2x√x) dx〗

Homework Equations



The Attempt at a Solution


∫▒〖(4x^2+2√x+1)/(2x√x) dx〗
∫▒((4x^2)/(2x^(3/2) )+(2√x)/(2x^(3/2) )+1/(2x^(3/2) ))dx
2∫▒x^(1/2) dx+∫▒〖x^(-1) dx〗+1/2 ∫▒x^(-3/2) dx
2×2/3 x^(3/2)+ln⁡x+1/2×(-2) x^(-1/2)+C
4/3 x^(3/2)+ln⁡x-x^(-1/2)+C
4/3 √(x^3 )+ln⁡x-1/√x+C
It looks good. Take the derivative to check it.
 
Thanks
 

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