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Have I understood this correctly? Integrating Ln[x] watch

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    Hello all!
    I am an upper 6th student doing maths and physics (not further maths) going on to do physics at Uni, and was trying to figure out how to integrate Ln[x], here's my method, could someone look over it and see if my reasoning and explanation of my working is clear, logical and actually possible?
    Thanks very much

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    Find:
     \int \ln (x) dx
    First, we must do integration by parts, using the formula:
     \int u \frac {dv} {dx} dx = uv - \int v \frac {du} {dx} dx
     u = \ln (v) where  v = x
    Find values of the derivative of u, with respect to x, and the differential of v with respect to x, as shown:
     \frac{dv}{dx}=1  dv = dx
     \frac{du}{dx} = \frac{1}{x}  du = \frac{dx}{x}
    Substitute in our values of u, dv, v and du, to get the following equation:
     \int \ln (x) dx = x \ln (x) - \int dx
    This makes the solution obvious (of course we must consider any unknown constants).
     \int \ln (x) dx = x \ln (x) - x + c
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    (Original post by timthegreat4)
    Hello all!
    I am an upper 6th student doing maths and physics (not further maths) going on to do physics at Uni, and was trying to figure out how to integrate Ln[x], here's my method, could someone look over it and see if my reasoning and explanation of my working is clear, logical and actually possible?
    Thanks very much

    Spoiler:
    Show
    Find:
     \int \ln (x) dx
    First, we must do integration by parts, using the formula:
     \int u \frac {dv} {dx} dx = uv - \int v \frac {du} {dx} dx
     u = \ln (x) = ln (v)  v = x
    Find values of the differential of u, with respect to x, and the differential of v with respect to x, as shown:
     \frac{dv}{dx}=1  dv = dx
     \frac{du}{dx} = \frac{1}{x}  du = \frac{dx}{x}
    Currently we are integrating everything with respect to x, but we should consider this equation as integrating with respect to v and u, as shown:
     \int udv = uv - \int vdu
    Substitute in our values of u, dv, v and du, to get the following equation:
     \int \ln (x) dx = x \ln (x) - \int dx
    This makes the solution obvious (of course we must consider any unknown constants).
     \int \ln (x) dx = x \ln (x) - x + c
    Your correct. The integral of lnx is xlnx - x.

    Just use integation by parts with:

    u = lnx
    v' = 1

    then apply integation by parts rules as usual.
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    (Original post by timthegreat4)
     u = \ln (x) = ln (v)  v = x
    Although your answer is correct, this line makes no sense.

    Also, you shouldn't be saying "differential", the correct word is "derivative".

    (Original post by timthegreat4)
    Currently we are integrating everything with respect to x, but we should consider this equation as integrating with respect to v and u, as shown:
     \int udv = uv - \int vdu
    Not really, no. Your first formula of integration by parts was more useful. "dv" and "du" don't really make sense by themselves - they're short forms for dv/dx or du/dx that people sometimes use, but you're not really integrating w.r.t u or v. So where you write stuff like

    "du = dx/x" - that doesn't strictly make sense either. It's just abusing notation because it still works.
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    (Original post by Swayum)
    Although your answer is correct, this line makes no sense.

    Also, you shouldn't be saying "differential", the correct word is "derivative".



    Not really, no. Your first formula of integration by parts was correct. "dv" and "du" don't really make sense by themselves - they're short forms for dv/dx or du/dx that people sometimes use, but you're not really integrating w.r.t u or v. So where you write stuff like

    "du = dx/x" - that doesn't strictly make sense either. It's just abusing notation because it still works.
    right, I fixed those few things u pointed out, thanks
    du = dx/x, you say it doesnt strictly work... however I can use that as a means for finding the integral of ln[x] correct?
    Thanks very much
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    (Original post by timthegreat4)
    right, I fixed those few things u pointed out, thanks
    du = dx/x, you say it doesnt strictly work... however I can use that as a means for finding the integral of ln[x] correct?
    Thanks very much
    I'm not saying it doesn't work - I'm saying that it's technically wrong to write that. It also doesn't help you to write it here anyway:

    The integration by parts formula start off with \int u\frac{\mathrm{d}v}{\mathrm{d}x} \mathrm{d}x

    So why would you bother rewriting dv/dx? You just literally sub it in there. So if dv/dx = sinx, you don't start wondering what "dv" is, you just sub it in there to get \int u\sin(x) \mathrm{d}x.

    Your adjustment is still illogical. Why are you saying u = ln(v)? I mean, it's true, but it doesn't make any sense to do that. I think your problem is that you're seeing integration by parts as the "dx" cancelling and so you think that we're integrating w.r.t v, so you need u as a function of v. This is not how to look at it. u is a function of x and so is dv/dx. Because they're both functions of x, we can integrate them w.r.t x. So your integrals will ALWAYS be w.r.t x and you shouldn't think of them as integrating w.r.t v or u.

    In other words, you should start by saying "Let u = ln(x) and let dv/dx = 1. Then the integration by parts formula tells me my integral is equivalent to uv - integral of vdu/dx"
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    yeah yeah I just realised what I did XD
    Its late lool, thanks
 
 
 
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