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    Integrate 1/(20-kV) dV

    u = 20-kV

    du/dV = -k

    dV = du/-k

    Therefore it becomes

    Integrate 1/-ku du

    = -k ln(20-kV)

    HOWEVER, the answer is actually \frac{1}{-k} ln(20-kV)

    How is this so? I'd appreciate somebody clarifying, thank you in advance.
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    (Original post by hopefuldentist10)
    Integrate 1/(20-kV) dV

    u = 20-kV

    du/dV = -k

    dV = du/-k

    Therefore it becomes

    Integrate 1/-ku du

    = -k ln(20-kV)

    HOWEVER, the answer is actually \frac{1}{-k} ln(20-kV)

    How is this so? I'd appreciate somebody clarifying, thank you in advance.
    Note that: \displaystyle\int -\dfrac{1}{ku} \ du = -\dfrac{1}{k} \int \dfrac{1}{u} \ du.
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    (Original post by hopefuldentist10)
    Integrate 1/(20-kV) dV

    u = 20-kV

    du/dV = -k

    dV = du/-k

    Therefore it becomes

    Integrate 1/-ku du

    = -k ln(20-kV)

    HOWEVER, the answer is actually \frac{1}{-k} ln(20-kV)

    How is this so? I'd appreciate somebody clarifying, thank you in advance.
    In this is instance, K just represents a constant. When you integrate a constant with respect to x, you make no changes to the constant itself.

    You can re-write the integral as I = -1/k • 1/u [du].

    So, integrating will give you -1/k • ln(u).

    Also, if you try differentiating your answer of -k • Ln(u) you will get -k/u and not 1/-ku.
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    Ahh, it was that simple. Thank you.
 
 
 
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